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Chapter 7  ·  Class XII Mathematics

Reversing the Derivative

Integrals

The Antidote to Differentiation — Standard Forms, Substitution, and Parts

📖 Read Full Notes ⚡ Formula Capsules

Chapter Snapshot

14Concepts
22Formulae
10–12%Exam Weight
4–6Avg Q's
HighDifficulty

Why This Chapter Matters for Entrance Exams

CBSEJEE MainJEE AdvancedBITSAT

Integrals is the highest-weightage chapter in Class XII Calculus for JEE — expect 4–6 questions across Main and Advanced combining substitution, partial fractions, and by-parts techniques. CBSE Boards allocate 10+ marks including at least one definite-integral-by-properties question.

Key Concept Highlights

Antiderivative / Indefinite Integral
Standard Integrals
Integration by Substitution
Integration Using Trigonometric Identities
Integrals of Particular Rational/Irrational Forms
Integration by Partial Fractions
Integration by Parts
ILATE Rule
Definite Integral as a Limit of a Sum
Fundamental Theorem of Calculus
Evaluation of Definite Integrals by Substitution
Properties of Definite Integrals
Integrals of Even/Odd Functions
Reduction Using King's Property

Important Formula Capsules

$\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C,\ n\ne -1$
$\int \dfrac{dx}{x^2+a^2} = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C$
$\int \dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\dfrac{x}{a} + C$
$\int \dfrac{dx}{\sqrt{x^2+a^2}} = \ln\left|x+\sqrt{x^2+a^2}\right| + C$
$\int u\,v\,dx = u\!\int v\,dx - \int\!\left(u'\!\int v\,dx\right)dx$
$\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx\ \text{(King's Property)}$
$\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx\ \text{if } f \text{ even},\ =0 \text{ if odd}$
$\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_0^a f(2a-x)\,dx$

What You Will Learn

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📖 Full Notes Complete NCERT Coverage 🧠 MCQ Practice Topic-wise Questions 🎯 PYQ Bank JEE · CBSE · BITSAT 📘✔️ Exercises Step by step Solutions to textbook exercises ✔️❌ True / False Concept-based True & False questions

🏆 Exam Strategy & Preparation Tips

King's Property (∫f(x)dx = ∫f(a+b−x)dx) single-handedly solves many "impossible-looking" definite integrals — always check for it first. Build fluency with all standard forms before attempting partial fractions or by-parts problems. This is the highest-yield chapter in the whole syllabus; allocate 3 weeks.

Chapter 7 · CBSE · Class XII
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Integrals

Integrals NCERT Class 12 Class 12 Mathematics Mathemaics Notes CBSE Class 12 NCERT Solutions Indefinite Integrals Definite Integrals Properties of Definite Integrals Fundamental Theorem of Integral Calculus Integration by Parts Integration by Substitution Standard Integrals Trigonometric Integrals Evaluation of Definite Integrals Area Function Even and Odd Functions Integration Formulas JEE Main JEE Advanced CUET Board Exam Competitive Exams
📋
Table of Contents
20 sections
📖 Introduction
Introduction to Integral Calculus
🗒️ Definition of Integration
Definition of Integration
Integration is the mathematical process of finding a function whose derivative is a given function. It is also used to calculate accumulated quantities, including areas under curves, through definite integrals.

Suppose that a function \(F(x)\) is differentiable and its derivative is \(f(x)\). Then,
\[ \frac{d}{dx}F(x)=f(x) \]

In this case, \(F(x)\) is called an antiderivative or primitive of \(f(x)\). The process of finding such an antiderivative is called integration.
Symbolically, integration is represented by
\[ \int\limits f(x)\,dx=F(x)+C \]

where:
  • \(\int\limits\) is the integral sign.
  • \(f(x)\) is the integrand, that is, the function to be integrated.
  • \(dx\) indicates the variable of integration.
  • \(F(x)\) is an antiderivative of \(f(x)\).
  • \(C\) is an arbitrary constant of integration.
The equation above represents an indefinite integral. It describes the entire family of antiderivatives of the integrand, rather than a single function.
🏛️ Historical Note
Historical Background and Development of Integral Calculus
The development of integral calculus arose from two important mathematical problems: finding a function when its derivative was known and determining areas bounded by curves. These problems eventually led to the development of indefinite and definite integrals.

Gottfried Wilhelm Leibniz (1646–1716) and Isaac Newton (1643–1727) independently developed fundamental ideas of calculus during the seventeenth century. Leibniz introduced much of the notation used in modern calculus, including the integral symbol \(\int\limits\), derived from an elongated form of the letter S, representing the Latin word *summa*, meaning sum.

The connection between differentiation and integration is expressed by the Fundamental Theorem of Calculus. This theorem establishes a relationship between antiderivatives and definite integrals, demonstrating that differentiation and integration are inverse processes under suitable conditions.
🗂️ The Two Fundamental Problems of Integral Calculus
Finding a Function When Its Derivative Is Given
The first problem is to determine an unknown function when its derivative is known. This is the reverse of differentiation.

For example, suppose
\[ \frac{dy}{dx}=2x \]
We need to find the function \(y\) whose derivative is \(2x\).

Since
\[ \frac{d}{dx}(x^2)=2x \]
we obtain
\[ y=x^2+C \]
where \(C\) is an arbitrary constant.

Thus, integration reverses differentiation by recovering a family of functions that possess the given derivative.

Finding the Area Bounded by a Curve
The second problem is to calculate the area of a region bounded by the graph of a function, the coordinate axes, and specified vertical lines.

Consider a continuous, non-negative function \(y=f(x)\) on the interval \([a,b]\). The area between its graph and the x-axis, from \(x=a\) to \(x=b\), is represented by the definite integral

\[ A=\int\limits_a^b f(x)\,dx \]
This formula applies directly when \(f(x)\geq 0\) throughout the interval. If the graph lies partly or entirely below the x-axis, the definite integral gives signed area; the geometric area must be calculated by accounting for the absolute value of the function or by splitting the interval appropriately.

The underlying idea is to divide the region into many narrow strips, approximate the area of each strip by a rectangle, and add these approximate areas. As the widths of the strips approach zero, the sum approaches the exact area under suitable conditions.
💡 Concept of Antiderivatives or Primitives
🤔 Did You Know?
Why Is the Constant of Integration Necessary?

Suppose two differentiable functions \(F(x)\) and \(G(x)\) have the same derivative on an interval. Then

\[ F'(x)=G'(x) \]
Rearranging,
\[ \frac{d}{dx}[F(x)-G(x)]=0 \]
A differentiable function whose derivative is zero throughout an interval is constant on that interval. Hence,
\[ F(x)-G(x)=C \]
or
\[ F(x)=G(x)+C \]
This proves that any two antiderivatives of the same function on an interval differ by a constant. Consequently, the arbitrary constant \(C\) must be included when evaluating an indefinite integral. Important note: Omitting \(+C\) from an indefinite integral is a common examination error. The constant is not required in the final evaluation of an ordinary definite integral because it cancels when the upper and lower limits are substituted into the same antiderivative.
🗒️ Indefinite Integrals
Definition: An indefinite integral is the family of all antiderivatives of a given function. It is written as

\[ \int\limits f(x)\,dx=F(x)+C \]
provided that \(F'(x)=f(x)\). An indefinite integral does not have upper and lower limits. Its result is a function together with an arbitrary constant, rather than a numerical value.
🗒️ Example
Evaluate \(\displaystyle\int 4x^3\,dx\)
Use the reverse of the power rule of differentiation.
  1. 1

    Identify the power of \(x\).

  2. 2

    Increase the power by one.

  3. 3

    Divide by the new power.

  4. 4

    Add the constant of integration.

Formula: 
\[\int\limits x^n\,dx=\frac{x^{n+1}}{n+1}+C, \qquad n\neq -1\]
  1. Using the power rule of integration, we get
    \[ \begin{aligned} \int\limits 4x^3\,dx &=4\int\limits x^3\,dx\\ &=4\left(\frac{x^4}{4}\right)+C\\ &=x^4+C \end{aligned} \]
Answer
\[\boxed{\int\limits 4x^3\,dx=x^4+C}\]
🛡️ Verification
✅
VerificationVerified
Differentiating the result gives
\[\frac{d}{dx}(x^4+C)=4x^3\]
which is the original integrand.
📘 Definite Integrals
🗒️ Example
Evaluate \(\displaystyle\int\limits_0^2 2x\,dx\).
Apply the Fundamental Theorem of Calculus.
  1. The antiderivative of \(2x\) is \(x^2\). Therefore,
    \[ \begin{aligned} \int\limits_0^2 2x\,dx &=[x^2]_0^2\\ &=2^2-0^2\\ &=4 \end{aligned} \]
Answer
\[\boxed{\int\limits_0^2 2x\,dx=4}\]
🔍
Geometrical interpretation
The graph \(y=2x\) is a straight line passing through the origin. Between \(x=0\) and \(x=2\), the region under the line is a right triangle with base \(2\) units and height \(4\) units. Its area is
\[ A=\frac{1}{2}\times 2\times 4=4 \]
This agrees with the value obtained using integration.
⚖️ Difference Between Indefinite and Definite Integrals
Basis Indefinite Integral Definite Integral
Notation \(\displaystyle\int f(x)\,dx\) \(\displaystyle\int\limits_a^b f(x)\,dx\)
Limits No upper or lower limits Has specified upper and lower limits
Result A family of antiderivatives A numerical value when the integral exists
Constant of integration Must be included as \(+C\) Not included in the final evaluated value
Primary interpretation Reconstruction of a function from its derivative Signed accumulation over an interval
Common application Finding an unknown function Finding area under a curve and accumulated quantities
🔗 Relationship Between Differentiation and Integration
Differentiation and integration are closely related mathematical operations. Differentiation determines the rate of change of a function, whereas integration reconstructs a function from its derivative or accumulates a quantity over an interval.

If
\[ \frac{d}{dx}F(x)=f(x) \]
then
\[ \int f(x)\,dx=F(x)+C \]
For example,
\[ \frac{d}{dx}(\sin x)=\cos x \]
Therefore,
\[ \int \cos x\,dx=\sin x+C \]
The reverse relationship must be applied carefully. Integration does not simply reverse differentiation by changing a symbol; the appropriate integration rule must be identified, and the constant of integration must be included for indefinite integrals.
🗂️ Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus connects indefinite and definite integration with differentiation. It is one of the most important results in integral calculus.
Differentiation of an Accumulation Function
Suppose \(f\) is continuous on an interval containing \(a\) and \(x\), and define
\[ G(x)=\int_a^x f(t)\,dt \]
Then
\[ G'(x)=f(x) \]
This result states that differentiating the accumulated integral recovers the original continuous function.
Evaluation of a Definite Integral
If \(F\) is an antiderivative of \(f\), so that \(F'(x)=f(x)\), and \(f\) is continuous on \([a,b]\), then
\[ \boxed{\int_a^b f(x)\,dx=F(b)-F(a)} \]
This result eliminates the need to calculate the area using infinitely many rectangles individually. Instead, one finds an antiderivative and evaluates it at the two limits.
🔢 Essential Formulae for Integration
🛠️ Mathematical Applications
  • Area calculation: Finding the area under a curve and the area enclosed between two curves.
  • Accumulation: Determining the total quantity accumulated from a known rate of change.
  • Motion: Finding displacement from velocity and changes in position from a known velocity function.
  • Geometry: Calculating areas of irregular regions that cannot be evaluated conveniently using elementary geometric formulae.
  • Physical sciences: Modelling work, mass distributions, charge distributions, and other quantities through suitable integrals.
  • Economics and statistics: Analysing accumulated costs, continuous probability distributions, and other continuous models.
🌟 Significance for CBSE Class 12 Board Examinations

Integrals are a central topic in Class 12 Mathematics and provide the foundation for subsequent applications of definite integration. A sound understanding of antiderivatives and standard formulae helps students solve textbook exercises, application-based questions, and multi-step problems.

Board examination preparation should focus on:

  • Understanding the meaning of an antiderivative and the role of the constant of integration.
  • Memorising standard integration formulae with their correct signs and restrictions.
  • Applying substitution, integration by parts, and partial fractions in the appropriate situations.
  • Evaluating definite integrals using the Fundamental Theorem of Calculus and the properties of definite integrals.
  • Writing complete mathematical steps, including transformations, substitutions, limits, and final answers.
  • Recognising the difference between signed area and geometrical area.
Significance for Competitive Entrance Examinations

Integration is an important component of the calculus syllabus for competitive examinations such as JEE Main and JEE Advanced. Questions may combine integration with trigonometric identities, algebraic simplification, logarithmic and exponential functions, definite integral properties, and graphical reasoning.

Competitive examination aspirants should develop the ability to:

  • Recognise integrands that can be simplified using algebraic or trigonometric identities.
  • Identify substitutions that transform complicated expressions into standard forms.
  • Choose efficiently between substitution, integration by parts, and partial fractions.
  • Use symmetry and other properties of definite integrals to reduce lengthy calculations.
  • Verify indefinite integral answers by differentiation.
  • Interpret definite integrals geometrically when the graph provides a simpler solution.

Preparation strategy: Build a strong foundation in differentiation before beginning advanced integration. Many integration techniques become easier when the student recognises the derivative of an inner function or the structure of a previously differentiated expression.

❌ Common Mistakes
  • Omitting the constant: Always include \(+C\) in an indefinite integral.
  • Misapplying the power rule: The formula \(\int x^n\,dx=x^{n+1}/(n+1)+C\) is not valid for \(n=-1\). Use \(\int 1/x\,dx=\ln|x|+C\) instead.
  • Sign errors: Remember that \(\int\sin x\,dx=-\cos x+C\), whereas \(\int\cos x\,dx=\sin x+C\).
  • Confusing definite and indefinite integrals: An indefinite integral gives a family of functions; a definite integral gives a value when it exists.
  • Incorrect limit substitution: In \([F(x)]_a^b\), calculate \(F(b)-F(a)\), not \(F(a)-F(b)\).
  • Ignoring the domain: Expressions involving logarithms, reciprocals, and inverse trigonometric functions may have domain restrictions.
  • Confusing signed and geometrical area: A definite integral can be negative if the graph lies below the x-axis.
  • Failing to verify: Differentiating an indefinite integral result is an effective way to detect errors.
📋 CBSE Case Study: HOTS-Based Application

Situation: A particle moves along a straight line, and its velocity at time \(t\) seconds is given by

\[ v(t)=4t+2 \]

where velocity is measured in metres per second.

Question 1: Find the displacement of the particle between \(t=0\) and \(t=3\) seconds, assuming the given velocity function holds throughout the interval.

Concept: Displacement is obtained by integrating velocity with respect to time.

\[ s(3)-s(0)=\int_0^3(4t+2)\,dt \]

\[ \begin{aligned} s(3)-s(0) &=[2t^2+2t]_0^3\\ &=(18+6)-0\\ &=24\text{ m} \end{aligned} \]

Answer: The displacement is \(24\) metres.

Question 2: Determine the antiderivative of the velocity function.

\[ \begin{aligned} s(t)&=\int(4t+2)\,dt\\ &=2t^2+2t+C \end{aligned} \]

Here, \(C\) represents the initial position when the position function is expressed in metres. If the initial position is specified, the constant can be determined.

Question 3: Why is the constant of integration absent from the final displacement calculation?

Answer: The constant cancels when the antiderivative is evaluated at the upper and lower limits:

\[ [F(t)+C]_0^3 =[F(3)+C]-[F(0)+C] =F(3)-F(0) \]

Higher-order thinking: Explain why integrating velocity gives displacement rather than necessarily giving the total distance travelled. The distinction is that displacement is the signed change in position, whereas total distance requires accounting for the magnitude of velocity over the interval.

🔑 Key Takeaway
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Integration as an Inverse Process of Differentiation

📋
Table of Contents
17 sections
📖 Introduction
📘 Definition of Integration as the Inverse Process
✏️ Understanding Integration Through Examples
1
Integration of a Trigonometric Function
We know that
\[\frac{d}{dx}(\sin x)=\cos x\]
Therefore, \(\sin x\) is an antiderivative of \(\cos x\).

However, the derivative of \(\sin x+3\) is also \(\cos x\), because the derivative of the constant \(3\) is zero. Similarly,
\[\frac{d}{dx}(\sin x-7)=\cos x\]
More generally, for every real constant \(C\),
\[\frac{d}{dx}(\sin x+C)=\cos x\]
Hence, the complete family of antiderivatives of \(\cos x\) is
\[\boxed{\int\cos x\,dx=\sin x+C}\]
where \(C\in\mathbb{R}\)

Conceptual conclusion: An indefinite integral represents infinitely many functions, all having the same derivative. The arbitrary constant distinguishes one member of this family from another.
2
Integration of a Polynomial Function
Since
\[\frac{d}{dx}(x^3)=3x^2,\]
we conclude that \(x^3\) is an antiderivative of \(3x^2\) Likewise,
\[\frac{d}{dx}(x^3+5)=3x^2\]
Therefore,
\[\boxed{\int 3x^2\,dx=x^3+C}\]
🛡️ Verification
✅
VerificationVerified
Differentiating the result gives
\[ \frac{d}{dx}(x^3+C)=3x^2\]
which is the original integrand.
✏️ Example
3
Question
Find all antiderivatives of \(6x^2\)
Reverse the power rule of differentiation and include the arbitrary constant.
  1. 1

    Identify a function whose derivative is \(6x^2\).

  2. 2

    Integrate using the power rule.

  3. 3

    Include the arbitrary constant.

  4. 4

    Verify the result by differentiation.

  1. \[ \begin{aligned} \int 6x^2\,dx &=6\int x^2\,dx\\ &=6\left(\frac{x^3}{3}\right)+C\\ &=2x^3+C. \end{aligned} \]
  2. Therefore, the required family of antiderivatives is
    \[\boxed{F(x)=2x^3+C,\quad C\in\mathbb{R}}\]
  3. verification
    \[ \frac{d}{dx}(2x^3+C)=6x^2\]
  4. Further interpretation: If an additional condition such as \(F(1)=4\) is given, the constant can be determined.
    \[2(1)^3+C=4\]
    \[C=2\]
  5. Hence, the particular antiderivative satisfying the condition is
    \[\boxed{F(x)=2x^3+2}\]
    This illustrates the distinction between a general antiderivative and a particular antiderivative determined by an initial condition.
🧮 Theorem
  1. Let \(g\) and \(h\) be differentiable functions on an interval \(I\) such that
    \[g'(x)=h'(x),\qquad \forall x\in I\]
  2. Define a new function \(u(x)\) by
    \[u(x)=g(x)-h(x)\]
  3. Differentiating both sides with respect to \(x\),
    \[\begin{aligned} u'(x) &=\frac{d}{dx}[g(x)-h(x)]\\ &=g'(x)-h'(x). \end{aligned}\]
  4. Since \(g'(x)=h'(x)\),
    \[u'(x)=0,\quad \forall x\in I.\]
  5. By the Mean Value Theorem, a differentiable function whose derivative is zero throughout an interval is constant on that interval. Therefore,
    \[u(x)=C\]
    for some real constant \(C\)
  6. Substituting \(u(x)=g(x)-h(x)\), we obtain
    \[\boxed{g(x)-h(x)=C}\]
  7. or equivalently,
    \[\boxed{g(x)=h(x)+C}\]
🧮 Theorem
Statement
If two differentiable functions \(g(x)\) and \(h(x)\) have equal derivatives throughout an interval \(I\), then they differ by a constant on that interval.

Mathematically, if
\[g'(x)=h'(x),\qquad \forall x\in I,\]
then there exists a constant \(C\) such that
\[\boxed{g(x)-h(x)=C,\qquad x\in I}\]
Proof
  1. Let \(g\) and \(h\) be differentiable functions on an interval \(I\) such that
    \[g'(x)=h'(x),\qquad \forall x\in I\]
  2. Define a new function \(u(x)\) by
    \[u(x)=g(x)-h(x)\]
  3. Differentiating both sides with respect to \(x\),
    \[\begin{aligned} u'(x) &=\frac{d}{dx}[g(x)-h(x)]\\ &=g'(x)-h'(x). \end{aligned}\]
  4. Since \(g'(x)=h'(x)\),
    \[u'(x)=0,\quad \forall x\in I.\]
  5. By the Mean Value Theorem, a differentiable function whose derivative is zero throughout an interval is constant on that interval. Therefore,
    \[u(x)=C\]
    for some real constant \(C\)
  6. Substituting \(u(x)=g(x)-h(x)\), we obtain
    \[\boxed{g(x)-h(x)=C}\]
  7. or equivalently,
    \[\boxed{g(x)=h(x)+C}\]
Hence proved
🌟 Significance of the theorem
This result establishes why the general indefinite integral contains the arbitrary constant. If \(F(x)\) and \(G(x)\) are both antiderivatives of the same function, they must differ only by a constant on the interval under consideration.

Mathematical condition: The statement is made on an interval. If the domain is disconnected, different constants may occur on different connected components.
🗒️ Relationship Between Derivatives and Integrals

Many standard integration formulae can be obtained by reversing familiar differentiation formulae. The following table presents the relationship between differentiation and integration.

Differentiation Corresponding Integration
\(\displaystyle \frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)=x^n,\ n\neq -1\) \(\displaystyle \int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\ n\neq -1\)
\(\displaystyle \frac{d}{dx}(\sin x)=\cos x\) \(\displaystyle \int\cos x\,dx=\sin x+C\)
\(\displaystyle \frac{d}{dx}(-\cos x)=\sin x\) \(\displaystyle \int\sin x\,dx=-\cos x+C\)
\(\displaystyle \frac{d}{dx}(\tan x)=\sec^2x\) \(\displaystyle \int\sec^2x\,dx=\tan x+C\)
\(\displaystyle \frac{d}{dx}(-\cot x)=\operatorname{cosec}^2x\) \(\displaystyle \int\operatorname{cosec}^2x\,dx=-\cot x+C\)
\(\displaystyle \frac{d}{dx}(\sec x)=\sec x\tan x\) \(\displaystyle \int\sec x\tan x\,dx=\sec x+C\)
\(\displaystyle \frac{d}{dx}(-\operatorname{cosec} x)=\operatorname{cosec} x\cot x\) \(\displaystyle \int\operatorname{cosec} x\cot x\,dx=-\operatorname{cosec} x+C\)
\(\displaystyle \frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}\) \(\displaystyle \int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x+C\)
\(\displaystyle \frac{d}{dx}(\cos^{-1}x)=-\frac{1}{\sqrt{1-x^2}}\) \(\displaystyle \int\frac{dx}{\sqrt{1-x^2}}=-\cos^{-1}x+C\)
\(\displaystyle \frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}\) \(\displaystyle \int\frac{dx}{1+x^2}=\tan^{-1}x+C\)
\(\displaystyle \frac{d}{dx}(e^x)=e^x\) \(\displaystyle \int e^x\,dx=e^x+C\)
\(\displaystyle \frac{d}{dx}(\ln|x|)=\frac{1}{x},\ x\neq0\) \(\displaystyle \int\frac{dx}{x}=\ln|x|+C\)
\(\displaystyle \frac{d}{dx}\left(\frac{a^x}{\ln a}\right)=a^x\) \(\displaystyle \int a^x\,dx=\frac{a^x}{\ln a}+C\)
In the last formula, \(a\) is a positive constant with \(a\neq1\). The trigonometric formulae apply wherever the functions involved are defined, and the inverse-trigonometric formulae use their standard real domains.

Notation note: The symbol \(\sin^{-1}x\) denotes the inverse sine function, not the reciprocal of \(\sin x\). Similarly, \(\tan^{-1}x\) denotes inverse tangent.
📐 Derivation of the Power Rule of Integration
📐 Derivation
Derivation of Selected Standard Integral Formulae
📐 Derivation
Integral of \(\cos x\)
Since
\[\frac{d}{dx}(\sin x)=\cos x\]
the reverse process gives
\[\boxed{\int\cos x\,dx=\sin x+C}\]
📐 Derivation
Integral of \(\sin x\)
Since
\[\frac{d}{dx}(-\cos x)=\sin x\]
we obtain
\[\boxed{\int\sin x\,dx=-\cos x+C}\]
Exam alert: The negative sign is essential. The derivative of \(\cos x\) is \(-\sin x\), not \(\sin x\).
📐 Derivation
Integral of \(\operatorname{cosec} x\cot x\)
We know that
\[\frac{d}{dx}(\operatorname{cosec} x)=-\operatorname{cosec} x\cot x\]
Multiplying by \(-1\),
\[\frac{d}{dx}(-\operatorname{cosec} x)=\operatorname{cosec} x\cot x\]
Hence,
\[\boxed{\int\operatorname{cosec} x\cot x\,dx=-\operatorname{cosec} x+C}\]
📐 Derivation
Integral of \(\dfrac{1}{\sqrt{1-x^2}}\)
Using the derivative of the inverse sine function,
\[ \frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}, \qquad -1<x<1, \]
we obtain
\[\boxed{\int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x+C}\]
Alternatively, since
\[\frac{d}{dx}(\cos^{-1}x)=-\frac{1}{\sqrt{1-x^2}}\]
we may write
\[\boxed{\int\frac{dx}{\sqrt{1-x^2}}=-\cos^{-1}x+C}\]
These expressions are equivalent because
\[ \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}, \qquad -1\leq x\leq1. \]
Therefore, their difference is a constant, as expected for two antiderivatives of the same function.
📐 Derivation
Integral of \(a^x\)
For \(a>0\) and \(a\neq1\), the derivative of \(a^x\) is
\[\frac{d}{dx}(a^x)=a^x\ln a\]
Consequently,
\[\frac{d}{dx}\left(\frac{a^x}{\ln a}\right)=a^x\]
Hence,
\[\boxed{\int a^x\,dx=\frac{a^x}{\ln a}+C}\]
For \(a=e\), since \(\ln e=1\), this reduces to
\[\int e^x\,dx=e^x+C\]
✏️ Example
Worked Examples with Solution Roadmaps
4
Question
Evaluate
\[\int(5x^4-3x^2+7)\,dx\]
Use the linearity of integration to integrate each term separately
  1. 1

    Split the polynomial into separate terms.

  2. 2

    Apply the power rule to the terms containing powers of \(x\).

  3. 3

    Integrate the constant term.

  4. 4

    Add one arbitrary constant to the complete result.

  1. \[ \begin{aligned} &\int(5x^4-3x^2+7)\,dx\\ &\quad=5\int x^4\,dx-3\int x^2\,dx+7\int dx\\ &\quad=5\left(\frac{x^5}{5}\right) -3\left(\frac{x^3}{3}\right)+7x+C\\ &\quad=x^5-x^3+7x+C. \end{aligned} \]
Integral of \(a^x\)
\[\boxed{x^5-x^3+7x+C}\]
✅
VerificationVerified
\[ \frac{d}{dx}(x^5-x^3+7x+C) =5x^4-3x^2+7 \]
5
Question
Find \(y\) if
\[\frac{dy}{dx}=4x^3\]
and \(y=6\) when \(x=1\).
Integrate the derivative to obtain the general solution, then use the given condition to determine the constant.
  1. Integrating both sides with respect to \(x\),
    \[ \begin{aligned} y&=\int4x^3\,dx\\ &=x^4+C \end{aligned} \]
  2. Using the condition \(y=6\) when \(x=1\),
    \[6=1^4+C\]
  3. Therefore,
    \[C=5\]
  4. Hence, the required function is
    \[\boxed{y=x^4+5}\]
🔍
Interpretation
The differential equation specifies the derivative but not the exact function. The additional condition selects one particular member of the family of antiderivatives.
6
Question
Show that \(\sin x\) and \(-\cos x+\dfrac{\pi}{2}\) are both antiderivatives of \(\cos x\).
  1. For the first function
    \[\frac{d}{dx}(\sin x)=\cos x\]
  2. For the second function,
    \[\frac{d}{dx}\left(-\cos x+\frac{\pi}{2}\right)=\sin x\]
    📝 The second derivative is \(\sin x\), not \(\cos x\). Therefore, the two proposed functions are not both antiderivatives of \(\cos x\).
  3. A correct pair is \(\sin x\) and \(\sin x+\dfrac{\pi}{2}\), because
    \[\frac{d}{dx}\left(\sin x+\frac{\pi}{2}\right)=\cos x\]
    📝 Thus, both are antiderivatives of \(\cos x\), and their difference is the constant \(\dfrac{\pi}{2}\).

    Learning point: Always verify an antiderivative by differentiation instead of relying only on the appearance of the expression.
🏷️ Important Properties of Indefinite Integration
Properties
Integration of a sum
\[ \int[f(x)+g(x)]\,dx =\int f(x)\,dx+\int g(x)\,dx. \]
Integration of a difference
\[ \int[f(x)-g(x)]\,dx =\int f(x)\,dx-\int g(x)\,dx. \]
Constant multiple rule
\[\int kf(x)\,dx=k\int f(x)\,dx\]
where \(k\) is a constant.
Verification by differentiation
If
\[\int f(x)\,dx=F(x)+C\]
then
\[\frac{d}{dx}[F(x)+C]=f(x)\]
❌ Common Mistakes
  • Omitting \(+C\): An indefinite integral represents a family of antiderivatives, so the constant of integration must be included.
  • Incorrect sign in the sine integral: The correct formula is \(\int\sin x\,dx=-\cos x+C\).
  • Incorrect inverse cosine formula: The correct expression is \(\int\dfrac{dx}{\sqrt{1-x^2}}=-\cos^{-1}x+C\), not \(+\cos^{-1}x+C\).
  • Incorrect cosecant derivative: The correct identity is \(\dfrac{d}{dx}(-\operatorname{cosec} x)=\operatorname{cosec} x\cot x\).
  • Using the power rule when \(n=-1\): The integral of \(x^{-1}\) is \(\ln|x|+C\).
  • Confusing inverse and reciprocal trigonometric functions: \(\sin^{-1}x\) means inverse sine, while \(\dfrac{1}{\sin x}=\operatorname{cosec} x\).
  • Assuming an initial condition is unnecessary: A derivative alone determines a family of functions; an additional condition is needed to determine the constant when a particular function is required.
  • Ignoring domain restrictions: The integrand and antiderivative must be considered on an appropriate interval where the expressions are defined.
⚡ Exam Tip
Exam Tips for CBSE Boards and Competitive Examinations
📋 HOTS Questions and Conceptual Practice
Question 1: Antiderivatives of a Function

Find the general antiderivative of \(f(x)=12x^3\).

Solution:

\[ \begin{aligned} \int12x^3\,dx &=12\left(\frac{x^4}{4}\right)+C\\ &=3x^4+C. \end{aligned} \]

Therefore,

\[ \boxed{F(x)=3x^4+C}. \]

Question 2: Equal Derivatives

If \(F'(x)=G'(x)\) throughout an interval, what can be concluded about \(F(x)\) and \(G(x)\)?

Answer: They differ by a constant on that interval:

\[ \boxed{F(x)-G(x)=C}. \]

Question 3: Determining a Constant

If \(F'(x)=2x\) and \(F(2)=7\), find \(F(x)\).

Solution:

\[ F(x)=\int2x\,dx=x^2+C. \]

Using \(F(2)=7\),

\[ 4+C=7\Rightarrow C=3. \]

Hence,

\[ \boxed{F(x)=x^2+3}. \]

Question 4: Identifying an Incorrect Formula

A student claims that

\[ \int\sin x\,dx=\cos x+C. \]

Is the statement correct? Explain.

Answer: No. Differentiating \(\cos x+C\) gives \(-\sin x\), not \(\sin x\). The correct formula is

\[ \boxed{\int\sin x\,dx=-\cos x+C}. \]

Question 5: Higher-Order Reasoning

Can two antiderivatives of the same function differ by a non-constant function on an interval?

Answer: No. If their derivatives are equal throughout the interval, the derivative of their difference is zero. By the Mean Value Theorem, their difference must be constant on that interval.

🔑 Key Takeaway
📌 Note
Frequently Asked Questions
∫

Some Properties of Indefinite Integrals

📋
Table of Contents
6 sections
🏷️ Properties
Indefinite integration follows several important mathematical properties that simplify the evaluation of integrals and provide a systematic approach to solving more complicated problems. These properties arise primarily from the relationship between differentiation and integration and from the fact that an indefinite integral represents a family of antiderivatives.
br> For NCERT Class 12 Mathematics, Chapter 7, Integrals, these properties form the foundation for integrating algebraic expressions, trigonometric functions, exponential functions, logarithmic functions, and combinations of different functions. They are particularly useful in board examinations and competitive entrance examinations, including JEE Main and JEE Advanced.
Properties
Property I: Differentiation and Integration Are Inverse Processes
Statement: Differentiation and indefinite integration are inverse operations in the following sense.

If \(f(x)\) is continuous on an interval and has an antiderivative, then
\[\boxed{\frac{d}{dx}\left[\int f(x)\,dx\right]=f(x)}\]
Conversely, if \(f(x)\) is differentiable on an interval, then \boxed{\int f'(x)\,dx=f(x)+C} where \(C\) is an arbitrary real constant called the constant of integration.

The first identity means that differentiating an indefinite integral recovers its integrand. The second identity means that integrating the derivative of a function recovers the original function up to an arbitrary constant.
Proof
Let \(F(x)\) be an antiderivative of \(f(x)\). By definition,
\[\frac{d}{dx}F(x)=f(x)\]
Therefore, the indefinite integral of \(f(x)\) is
\[\int f(x)\,dx=F(x)+C\]
Differentiating both sides,
\[ \begin{aligned} \frac{d}{dx}\left[\int f(x)\,dx\right] &=\frac{d}{dx}[F(x)+C]\\ &=F'(x)+0\\ &=f(x) \end{aligned} \]
Hence,
\[\boxed{\frac{d}{dx}\left[\int f(x)\,dx\right]=f(x)}\]
For the converse, by the definition of the derivative,
\[f'(x)=\frac{d}{dx}f(x)\]
Integrating both sides with respect to \(x\),
\[ \int f'(x)\,dx =\int\frac{d}{dx}f(x)\,dx \]
Integration reverses differentiation, giving
\[\boxed{\int f'(x)\,dx=f(x)+C}\]
Hence proved
Verification of Property I
Question: Verify that differentiation and integration are inverse operations for \(f(x)=3x^2+2x\).

Solution: First, integrate the function:
\[ \begin{aligned} \int(3x^2+2x)\,dx &=3\int x^2\,dx+2\int x\,dx\\ &=x^3+x^2+C \end{aligned} \]
Now differentiate the result:
\[ \begin{aligned} \frac{d}{dx}(x^3+x^2+C) &=3x^2+2x\\ &=f(x) \end{aligned} \]
Thus, the original integrand is recovered, verifying the property.

Final result:
\[\boxed{\frac{d}{dx}\left[\int(3x^2+2x)\,dx\right]=3x^2+2x}\]
Property II: Indefinite Integrals with the Same Derivative Are Equivalent
Statement: If two differentiable functions have the same derivative throughout an interval, then they differ by a constant on that interval. Consequently, they represent the same family of antiderivatives. Suppose \(F(x)\) and \(G(x)\) are two antiderivatives of the same function \(f(x)\). Then
\[F'(x)=G'(x)=f(x)\]
It follows that
\[\boxed{F(x)-G(x)=C}\]
for some real constant \(C\)
Proof
Let \(F(x)\) and \(G(x)\) be differentiable functions on an interval \(I\), and suppose that
\[F'(x)=G'(x),\qquad x\in I\]
Consider the function
\[H(x)=F(x)-G(x)\]
Differentiating,
\[ \begin{aligned} H'(x) &=F'(x)-G'(x)\ &=0. \end{aligned} \]
By the Mean Value Theorem, a differentiable function whose derivative is zero throughout an interval is constant on that interval. Hence,
\[H(x)=C\]
Substituting \(H(x)=F(x)-G(x)\), we obtain
\[\boxed{F(x)-G(x)=C}\]
Therefore,
\[\boxed{F(x)=G(x)+C}\]
Hence proved.
Interpretation in Terms of Indefinite Integrals
Suppose \(f(x)\) and \(g(x)\) have the same derivative:
\[f'(x)=g'(x)\]
Then
\[f(x)=g(x)+C\]
Integrating, we obtain
\[\int f(x)\,dx=\int g(x)\,dx+C\]
Thus, the two indefinite integrals represent the same family of antiderivatives. The constant can be absorbed into the arbitrary constant already present in an indefinite integral. More precisely, if \(F'(x)=G'(x)\), then
\[ \{F(x)+C_1:C_1\in\mathbb R\} = \{G(x)+C_2:C_2\in\mathbb R\} \]
Important clarification: Two indefinite integrals are equivalent in this sense when their integrands are equal, or equivalently when their antiderivatives differ only by a constant. Two arbitrary integrals with different integrands are not necessarily equivalent.
Example 2: Identifying Equivalent Antiderivatives
Question: Show that \(x^2+3\) and \(x^2-5\) are antiderivatives of the same function.

Solution: Differentiating the first function,
\[\frac{d}{dx}(x^2+3)=2x\]
Differentiating the second function,
\[\frac{d}{dx}(x^2-5)=2x\]
Both functions have the same derivative. Their difference is
\[(x^2+3)-(x^2-5)=8\]
Since the difference is constant, both belong to the same family of antiderivatives:
\[\boxed{\int 2x\,dx=x^2+C}\]
Property III: Integral of the Sum of Two Functions
Statement: The indefinite integral of the sum of two integrable functions is equal to the sum of their indefinite integrals, up to an arbitrary constant.
\[\boxed{\int[f(x)+g(x)]\,dx=\int f(x)\,dx+\int g(x)\,dx}\]
This property is called the sum rule of integration. It allows an integrand containing several terms to be separated into simpler integrals.
Proof
Let
\[F(x)=\int f(x)\,dx\]
and
\[G(x)=\int g(x)\,dx\]
By Property I,
\[F'(x)=f(x),\qquad G'(x)=g(x)\]
Using the linearity of differentiation,
\[ \begin{aligned} \frac{d}{dx}[F(x)+G(x)] &=F'(x)+G'(x)\\ &=f(x)+g(x) \end{aligned} \]
Therefore, \(F(x)+G(x)\) is an antiderivative of \(f(x)+g(x)\). Hence,
\[\boxed{\int[f(x)+g(x)]\,dx=\int f(x)\,dx+\int g(x)\,dx}\]
where the arbitrary constant on the right-hand side is understood to be combined into a single constant.
Example 3: Integration of a Sum
Question: Evaluate
\[\int(4x^3+6x+5)\,dx\]
Concept: Separate the polynomial into individual terms and integrate each term independently. Solution roadmap:
  1. Apply the sum rule.
  2. Use the power rule for \(x^3\) and \(x\).
  3. Integrate the constant term.
  4. Add the constant of integration.
Step-by-step Solution:
\[ \begin{aligned} &\int(4x^3+6x+5)\,dx\\ &\quad=4\int x^3\,dx+6\int x\,dx+5\int dx\\ &\quad=4\left(\frac{x^4}{4}\right) +6\left(\frac{x^2}{2}\right)+5x+C\\ &\quad=x^4+3x^2+5x+C \end{aligned} \]
Answer:
\[\boxed{\int(4x^3+6x+5)\,dx=x^4+3x^2+5x+C}\]
Property IV: Integral of a Constant Multiple of a Function
Statement: If \(k\) is a real constant and \(f(x)\) is integrable, then
\[\boxed{\int kf(x)\,dx=k\int f(x)\,dx}\]
This property is known as the constant multiple rule of integration. It allows a constant factor to be taken outside the integral sign.
Proof
Let
\[F(x)=\int f(x)\,dx\]
By Property I,
\[F'(x)=f(x)\]
Using the constant multiple rule of differentiation,
\[\begin{aligned}\frac{d}{dx}[kF(x)]&=kF'(x)\\&=kf(x).\end{aligned}\]
Thus, \(kF(x)\) is an antiderivative of \(kf(x)\). Consequently,
\[\boxed{\int kf(x)\,dx=k\int f(x)\,dx}\]
with the arbitrary constant understood appropriately.

Special case: When \(k=0\), the integral of the zero function is a constant:
\[\int 0\,dx=C\]
Example 4: Taking a Constant Outside the Integral
Question: Evaluate
\[\int 7\sin x\,dx\]
Solution:

Taking the constant \(7\) outside the integral,
\[ \begin{aligned} \int7\sin x\,dx &=7\int\sin x\,dx\\ &=7(-\cos x)+C \end{aligned} \]
Therefore,
\[\boxed{\int7\sin x\,dx=-7\cos x+C}\]
Verification:
\[\frac{d}{dx}(-7\cos x+C)=7\sin x\]
Property V: Generalisation to a Finite Number of Functions
Statement: The sum rule and constant multiple rule can be combined and extended to any finite number of integrable functions. Let \(f_1(x),f_2(x),\ldots,f_n(x)\) be integrable functions, and let \(k_1,k_2,\ldots,k_n\) be real constants. Then
\[ \boxed{ \begin{aligned} &\int\left[ k_1f_1(x)+k_2f_2(x)+\cdots+k_nf_n(x) \right]dx\\ &\quad=k_1\int f_1(x)\,dx +k_2\int f_2(x)\,dx +\cdots+k_n\int f_n(x)\,dx \end{aligned} } \]
This is the linearity property of indefinite integration.
Proof
Let \(F_i(x)\) be an antiderivative of \(f_i(x)\), for each \(i=1,2,\ldots,n\). Thus,
\[F_i'(x)=f_i(x)\]
Consider the function
\[H(x)=k_1F_1(x)+k_2F_2(x)+\cdots+k_nF_n(x)\]
Differentiating term by term,
\[ \begin{aligned} H'(x) &=k_1F_1'(x)+k_2F_2'(x)+\cdots+k_nF_n'(x)\\ &=k_1f_1(x)+k_2f_2(x)+\cdots+k_nf_n(x). \end{aligned} \]
Therefore, \(H(x)\) is an antiderivative of the combined integrand. Hence,
\[ \boxed{ \begin{aligned} &\int\left[\sum_{i=1}^{n}k_if_i(x)\right]dx\\ &\quad=\sum_{i=1}^{n}k_i\int f_i(x)\,dx. \end{aligned} } \]
Hence proved
Example 5: Applying the Generalised Linearity Property
Question: Evaluate
\[\int(6x^2-4\cos x+3e^x)\,dx\]
Solution: Using the linearity property,
\[ \begin{aligned} &\int(6x^2-4\cos x+3e^x)\,dx\\ &\quad=6\int x^2\,dx-4\int\cos x\,dx +3\int e^x\,dx\\ &\quad=6\left(\frac{x^3}{3}\right) -4\sin x+3e^x+C\\ &\quad=2x^3-4\sin x+3e^x+C \end{aligned} \]
Answer:
\[ \boxed{ \int(6x^2-4\cos x+3e^x)\,dx =2x^3-4\sin x+3e^x+C } \]
Additional Important Consequences of These Properties
Integral of a Difference
By applying the sum rule and the constant multiple rule with \(k=-1\), we obtain
\[\boxed{\int[f(x)-g(x)]\,dx=\int f(x)\,dx-\int g(x)\,dx}\]
For example,
\[ \begin{aligned} \int(5x^4-2x)\,dx &=5\int x^4\,dx-2\int x\,dx\\ &=x^5-x^2+C \end{aligned} \]
Integral of a Constant Function
If \(k\) is a constant, then
\[\boxed{\int k\,dx=kx+C}\]
This follows because
\[\frac{d}{dx}(kx+C)=k\]
A Constant Can Be Absorbed into the Integration Constant
Suppose two terms produce arbitrary constants \(C_1\) and \(C_2\). Their sum or difference is still an arbitrary constant:
\[C_1+C_2=C,\quad C_1-C_2=C\]
Therefore, when adding several indefinite integrals, it is sufficient to write one arbitrary constant in the final answer.
Important Distinction: What These Properties Do Not Permit
The linearity properties allow sums, differences, and constant multiples to be separated. However, they do not allow the integral of a product or quotient to be separated in the same way.

In general,
\[\boxed{\int f(x)g(x)\,dx\neq\left(\int f(x)\,dx\right)\left(\int g(x)\,dx\right)}\]
Similarly,
\[\boxed{\int\frac{f(x)}{g(x)}\,dx\neq\frac{\int f(x)\,dx}{\int g(x)\,dx}}\]
\[\boxed{\int\frac{f(x)}{g(x)}\,dx\neq\frac{\int f(x)\,dx}{\int g(x)\,dx}}\]
Illustration: Consider
\[\int x\cdot x\,dx=\int x^2\,dx=\frac{x^3}{3}+C\]
Separately integrating each factor and multiplying would give
\[\left(\int x\,dx\right)\left(\int x\,dx\right)=\left(\frac{x^2}{2}+C_1\right)\left(\frac{x^2}{2}+C_2\right),\]
which is not the antiderivative of \(x^2\). This demonstrates why integration is linear but is not generally multiplicative.
❌ Common Mistakes
  • Omitting the constant of integration: Write \(+C\) when presenting the final indefinite integral.
  • Writing an incorrect proof: When proving an integration property, differentiate both proposed antiderivatives and use the fact that functions with equal derivatives on an interval differ by a constant.
  • Confusing derivatives with integrals: The derivative of an indefinite integral recovers the integrand, whereas integrating a derivative recovers the original function plus a constant.
  • Applying linearity incorrectly: Constants may be taken outside the integral, and sums or differences may be separated, but products and quotients cannot generally be separated.
  • Writing several unnecessary constants: Combine all arbitrary constants into a single \(C\) in the final result.
  • Ignoring the interval: The theorem that two functions with equal derivatives differ by a constant applies on an interval. On disconnected domains, the constants may differ between connected components.
  • Forgetting negative coefficients: A minus sign outside an integral must remain attached to the corresponding integrated term.
⚡ Exam Tip
Exam Tips for CBSE Class 12 and Competitive Examinations
📋 HOTS and Application-Based Questions
Question 1: Use of Linearity

Evaluate

\[ \int(8x^3+6x^2-4)\,dx. \]

Solution:

\[ \begin{aligned} \int(8x^3+6x^2-4)\,dx &=8\int x^3\,dx+6\int x^2\,dx-4\int dx\\ &=2x^4+2x^3-4x+C. \end{aligned} \]

Hence,

\[ \boxed{2x^4+2x^3-4x+C} \]

Question 2: Verify an Integration Property

Show that

\[ \int[2f(x)+3g(x)]\,dx =2\int f(x)\,dx+3\int g(x)\,dx. \]

Solution: Let \(F'(x)=f(x)\) and \(G'(x)=g(x)\). Then

\[ \frac{d}{dx}[2F(x)+3G(x)] =2f(x)+3g(x). \]

Therefore, \(2F(x)+3G(x)\) is an antiderivative of \(2f(x)+3g(x)\). Hence,

\[ \boxed{ \int[2f(x)+3g(x)]\,dx =2\int f(x)\,dx+3\int g(x)\,dx } \]

Question 3: Identify the Incorrect Statement

A student claims that

\[ \int[f(x)+g(x)]\,dx =\int f(x)\,dx\cdot\int g(x)\,dx. \]

Is the claim correct?

Answer: No. The sum rule states that

\[ \boxed{ \int[f(x)+g(x)]\,dx =\int f(x)\,dx+\int g(x)\,dx } \]

The integral of a sum is the sum of the integrals, not their product.

Question 4: Higher-Order Reasoning

If \(F'(x)=G'(x)\) and \(F(0)=2\), \(G(0)=7\), determine \(F(x)-G(x)\).

Solution: Since the derivatives are equal, their difference is constant:

\[ F(x)-G(x)=C \]

Using \(x=0\),

\[ C=F(0)-G(0)=2-7=-5 \]

Therefore,

\[ \boxed{F(x)-G(x)=-5} \]

🔑 Key Takeaway
📌 Note
Frequently Asked Questions
∫

Example 1

📋
Table of Contents
4 sections
❓ Question
Finding Antiderivatives by the Method of Inspection
Write an antiderivative for each of the following functions using the method of inspection.
  1. \(\cos 2x\)
  2. \(3x^2+4x^3\)
  3. \(\dfrac{1}{x},\quad x\neq0\)
💡 Concept
🧩 Solution
Part (i)
\(\cos 2x\)
  1. We know that
    \[\frac{d}{dx}(\sin 2x)=2\cos 2x\]
  2. Dividing both sides by \(2\),
    \[\cos 2x=\frac{1}{2}\frac{d}{dx}(\sin 2x)\]
  3. Therefore,
    \[\cos 2x=\frac{d}{dx}\left(\frac{1}{2}\sin 2x\right)\]
  4. Hence, an antiderivative of \(\cos 2x\) is
    \[\boxed{\frac{1}{2}\sin 2x}\]
  5. The general indefinite integral is
    \[\boxed{\int\cos 2x\,dx=\frac{1}{2}\sin 2x+C}\]
  6. Verification
    \[ \frac{d}{dx}\left(\frac{1}{2}\sin 2x+C\right) =\frac{1}{2}(2\cos 2x)=\cos 2x. \]
  7. <strong>Exam tip:</strong> Whenever the argument of a trigonometric function is \(ax+b\), check the derivative of the inner function. For example,
    \[ \int\cos(ax+b)\,dx =\frac{1}{a}\sin(ax+b)+C,\quad a\neq0. \]
Part (ii)
\(3x^2+4x^3\)
  1. We know that
    \[\frac{d}{dx}(x^3)=3x^2\]
    and
    \[\frac{d}{dx}(x^4)=4x^3\]
  2. Adding these equations,
    \[\frac{d}{dx}(x^3+x^4)=3x^2+4x^3\]
  3. Therefore, an antiderivative of \(3x^2+4x^3\) is
    \[\boxed{x^3+x^4}\]
  4. The general indefinite integral is
    \[\boxed{\int(3x^2+4x^3)\,dx=x^3+x^4+C}\]
  5. Verification:
    \[\frac{d}{dx}(x^3+x^4+C)=3x^2+4x^3\]
  6. Exam tip: For polynomial expressions, inspect each term separately. The power rule can often identify the antiderivative immediately:
    \[\int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\quad n\neq-1\]
Part (iii)
\(\dfrac{1}{x},\quad x\neq0\)
  1. For \(x>0\),
    \[\frac{d}{dx}(\ln x)=\frac{1}{x}\]
  2. For \(x<0\), we use \(\ln(-x)\). By the chain rule,
    \[ \begin{aligned} \frac{d}{dx}\ln(-x) &=\frac{1}{-x}\cdot(-1)\\ &=\frac{1}{x} \end{aligned} \]
    Thus, the expression \(\ln|x|\) has derivative \(1/x\) for every \(x\neq0\).
  3. Therefore, an antiderivative is
    \[\boxed{\ln|x|}\]
  4. The general indefinite integral is
    \[\boxed{\int\frac{1}{x}\,dx=\ln|x|+C,\quad x\neq0}\]
  5. Verification:
    \[\frac{d}{dx}\ln|x|=\frac{1}{x},\quad x\neq0\]
  6. Important note: The absolute value is necessary when the domain includes negative values of \(x\). Since \(1/x\) is undefined at \(x=0\), the antiderivative is considered separately on the intervals \(x>0\) and \(x<0\). The integration constants on these two intervals need not be equal.
🔑 Key Takeaway
∫

Example 2

📋
Table of Contents
5 sections
❓ Question
Find the following integrals:
  1. \(\int\frac{x^3-1}{x^2}\,dx\)
  2. \(\int\left(x^{\frac23}+1\right)dx\)
🗺️ Roadmap

  1. Separate the numerator into individual terms.

    Simplify each term using the laws of exponents.

    Apply the linearity property of integration.

    Use the power rule and add the constant of integration.

🧩 Solution
Part (i)
\[\int\frac{x^3-1}{x^2}\,dx\]
  1. Split the fraction
    \[ \begin{aligned} \int\frac{x^3-1}{x^2}\,dx &=\int\left(\frac{x^3}{x^2}-\frac{1}{x^2}\right)dx \end{aligned} \]
  2. Simplify using the laws of exponents.
    Since \(\dfrac{x^m}{x^n}=x^{m-n}\) for \(x\neq0\),
    \[ \begin{aligned} \int\frac{x^3-1}{x^2}\,dx &=\int\left(x-x^{-2}\right)dx \end{aligned} \]
  3. Apply the linearity property.
    \[ \begin{aligned} \int\left(x-x^{-2}\right)dx &=\int x\,dx-\int x^{-2}\,dx \end{aligned} \]
  4. ⭐ The power rule states that
    \[ \int x^n\,dx=\frac{x^{n+1}}{n+1}+C, \qquad n\neq-1. \]
  5. Apply the power rule of integration
    \[ \begin{aligned} \int x\,dx-\int x^{-2}\,dx &=\frac{x^2}{2}-\frac{x^{-1}}{-1}+C\\ &=\frac{x^2}{2}+x^{-1}+C\\ &=\frac{x^2}{2}+\frac{1}{x}+C \end{aligned} \]
  6. Verification by Differentiation
    \[ \begin{aligned} \frac{d}{dx}\left(\frac{x^2}{2}+\frac{1}{x}+C\right) &=x-\frac{1}{x^2}\\ &=\frac{x^3-1}{x^2} \end{aligned} \]
    📝 The derivative equals the original integrand. Hence, the result is verified
Part (ii)
\[\int\left(x^{\frac23}+1\right)dx\]
  1. Apply the sum rule
    \[\begin{aligned}\int\left(x^{\frac23}+1\right)dx&=\int x^{\frac23}\,dx+\int1\,dx\end{aligned}\]
  2. Apply the power rule.
    \[ \begin{aligned} \int x^{\frac23}\,dx &=\frac{x^{\frac23+1}}{\frac23+1}\\ &=\frac{x^{\frac53}}{\frac53}\\ &=\frac35x^{\frac53} \end{aligned} \]
  3. Integrate the constant function
    \[\int1\,dx=x\]
  4. Combine the results
    \[ \begin{aligned} \int\left(x^{\frac23}+1\right)dx &=\frac35x^{\frac53}+x+C \end{aligned} \]
  5. Verification by Differentiation
    \[ \begin{aligned} \frac{d}{dx}\left(\frac35x^{\frac53}+x+C\right) &=\frac35\cdot\frac53x^{\frac23}+1\\ &=x^{\frac23}+1 \end{aligned} \]
    📝 This matches the original integrand, verifying the answer.
Answer
\[ \boxed{ \int\left(x^{\frac23}+1\right)dx =\frac35x^{\frac53}+x+C } \]
⚡ Exam Tip
❌ Common Mistakes
  • Writing \(\int x^{-2}\,dx=-x^{-1}\) without considering the negative sign introduced by the power rule. The correct result is \(-x^{-1}+C\).
  • Forgetting that subtracting the integral of \(x^{-2}\) produces a positive term \(1/x\) in Example (i).
  • Adding \(1\) instead of \(x\) when integrating the constant function \(1\).
  • Dividing by the original fractional exponent rather than by the exponent obtained after adding \(1\).
  • Omitting the constant of integration from the final result.
∫

Example 3

📋
Table of Contents
3 sections
❓ Question
Find the following integrals:
  1. \(\displaystyle\int(\sin x+\cos x)\,dx\)
  2. \(\displaystyle\int\operatorname{cosec} x(\operatorname{cosec} x+\cot x)\,dx\)
  3. \(\displaystyle\int\frac{1-\sin x}{\cos^2x}\,dx\)
🧩 Solution
Part (i)
\(\displaystyle\int(\sin x+\cos x)\,dx\)
  1. 💡
    Concept
    Concept used: Apply the sum rule of integration and the standard integrals of sine and cosine.
    🗺️ Roadmap
    1

    Separate the sum into two integrals.

    2

    Integrate each trigonometric function using standard formulae.

    3

    Combine the results and include the constant of integration.

  2. Using the sum rule,
    \[ \begin{aligned} \int(\sin x+\cos x)\,dx &=\int\sin x\,dx+\int\cos x\,dx \end{aligned} \]
  3. We know that
    \[\int\sin x\,dx=-\cos x+C\]
    and
    \[\int\cos x\,dx=\sin x+C\]
  4. Therefore, combining the results and absorbing the constants into a single constant,
    \[ \begin{aligned} \int(\sin x+\cos x)\,dx &=-\cos x+\sin x+C. \end{aligned} \]
  5. Verification:
    \[ \frac{d}{dx}(\sin x-\cos x+C) =\cos x+\sin x \]
    📝 This is the original integrand, so the result is correct
Answer
\[\boxed{\int(\sin x+\cos x)\,dx=\sin x-\cos x+C}\]
Part (ii)
\(\displaystyle\int\operatorname{cosec} x(\operatorname{cosec} x+\cot x)\,dx\)
  1. 💡
    Concept
    Expand the product and use the standard integrals of \(\operatorname{cosec}^2x\) and \(\operatorname{cosec} x\cot x\).
    🗺️ Roadmap
    1

    Expand the product inside the integral.

    2

    Separate the resulting terms using the sum rule.

    3

    Apply the standard trigonometric integration formulae, paying attention to signs.

    4

    Combine the results and verify by differentiation.

  2. Expanding the product,
    \[ \begin{aligned} &\int\operatorname{cosec} x(\operatorname{cosec} x+\cot x)\,dx\\ &\quad=\int(\operatorname{cosec}^2x+\operatorname{cosec} x\cot x)\,dx \end{aligned} \]
  3. Applying the sum rule,
    \[ \begin{aligned} &\int(\operatorname{cosec}^2x+\operatorname{cosec} x\cot x)\,dx\\ &\quad=\int\operatorname{cosec}^2x\,dx +\int\operatorname{cosec} x\cot x\,dx \end{aligned} \]
  4. ⭐ Using the standard formulae,
    \[\int\operatorname{cosec}^2x\,dx=-\cot x+C\]
    and
    \[\int\operatorname{cosec} x\cot x\,dx=-\operatorname{cosec} x+C\]
  5. Hence,
    \[ \begin{aligned} &\int\operatorname{cosec} x(\operatorname{cosec} x+\cot x)\,dx\\ &\quad=-\cot x-\operatorname{cosec} x+C \end{aligned} \]
  6. Verification:
    \[ \begin{aligned} \frac{d}{dx}(-\cot x-\operatorname{cosec} x+C) &=\operatorname{cosec}^2x+\operatorname{cosec} x\cot x\\ &=\operatorname{cosec} x(\operatorname{cosec} x+\cot x) \end{aligned} \]
    📝 Thus, the result is verified.
Answer
\[\boxed{\int\operatorname{cosec} x(\operatorname{cosec} x+\cot x)\,dx=-\cot x-\operatorname{cosec} x+C}\]
Part (iii)
\(\displaystyle\int\frac{1-\sin x}{\cos^2x}\,dx\)
  1. 💡
    Concept
    Split the fraction into simpler trigonometric terms and rewrite them using \(\sec x\), \(\tan x\), and \(\cot x\).
    🗺️ Roadmap
    1

    Separate the numerator into two terms.

    2

    Rewrite the first term as \(\sec^2x\).

    3

    Rewrite the second term using \(\tan x\sec x\).

    4

    Apply the standard integration formulae and verify the result.

  2. Split the fraction
    \[ \begin{aligned} \int\frac{1-\sin x}{\cos^2x}\,dx &=\int\left(\frac{1}{\cos^2x} -\frac{\sin x}{\cos^2x}\right)dx. \end{aligned} \]
  3. Express the terms using standard trigonometric ratios.
    Since
    \[\frac{1}{\cos^2x}=\sec^2x\]
    and
    \[\frac{\sin x}{\cos^2x}=\frac{\sin x}{\cos x}\cdot\frac{1}{\cos x}=\tan x\sec x,\]
  4. we obtain
    \[ \begin{aligned} \int\frac{1-\sin x}{\cos^2x}\,dx &=\int(\sec^2x-\tan x\sec x)\,dx\\ &=\int\sec^2x\,dx-\int\sec x\tan x\,dx \end{aligned} \]
  5. ⭐ Apply standard integration formulae.
    \[\int\sec^2x\,dx=\tan x+C\]
    and
    \[\int\sec x\tan x\,dx=\sec x+C\]
  6. Therefore,
    \[\begin{aligned}\int\frac{1-\sin x}{\cos^2x}\,dx&=\tan x-\sec x+C\end{aligned}\]
  7. Verification:
    \[ \begin{aligned} \frac{d}{dx}(\tan x-\sec x+C) &=\sec^2x-\sec x\tan x\\ &=\frac{1}{\cos^2x} -\frac{\sin x}{\cos^2x}\\ &=\frac{1-\sin x}{\cos^2x} \end{aligned} \]
    📝 Hence, the answer is verified.
Answer
\[\boxed{\int\frac{1-\sin x}{\cos^2x}\,dx=\tan x-\sec x+C}\]
🔑 Key Takeaway
∫

Methods of Integration

📋
Table of Contents
13 sections
🗺️ Overview

Integration is the inverse process of differentiation. However, unlike differentiation, which follows a set of systematic rules, integration often requires selecting an appropriate technique according to the structure of the integrand. The objective is to transform a complicated integral into a simpler form that can be evaluated using standard integration formulas.

In Class 12 Mathematics, the three important methods of integration are:

  1. Integration by Substitution: Used when a suitable change of variable simplifies the integrand, especially when the derivative of an inner function is also present.
  2. Integration Using Partial Fractions: Used mainly for rational functions that can be expressed as the quotient of two polynomials, provided the denominator can be factorised appropriately.
  3. Integration by Parts: Used for integrals involving products of functions when one factor becomes simpler upon differentiation and the other can be integrated.

The choice of method depends on the mathematical structure of the integrand. Recognising this structure is an important skill for solving NCERT exercises, CBSE board questions and competitive entrance examination problems.

📘 Integration by Substitution
💡 Basic Concept
🤔 Did You Know?
Why Does Substitution Work?
The method follows directly from the chain rule. If \(F\) is an antiderivative of \(f\), then
\[F'(x)=f(x)\]
After substituting \(x=g(t)\), the chain rule gives
\[ \frac{d}{dt}F(g(t)) =F'(g(t))g'(t) \]
Since \(F'(g(t))=f(g(t))\), we have
\[\frac{d}{dt}F(g(t))=f(g(t))g'(t)\]
Integrating with respect to \(t\), we obtain
\[\int f(g(t))g'(t)\,dt=F(g(t))+C\]
This demonstrates that substitution reverses the chain rule of differentiation. It allows us to recognise an integral as the derivative of a composite function and recover its antiderivative.
🧰 How to Choose a Suitable Substitution
The main challenge in this method is identifying a substitution that simplifies the integrand. A useful strategy is to look for an inner function whose derivative also occurs in the integrand.
  • Expression raised to a power: For an expression such as \((3x+2)^5\), consider \(t=3x+2\).
  • Function inside another function: For an expression such as \(\sin(x^2)\), consider \(t=x^2\) if the remaining factors support this choice.
  • Denominator and its derivative: For an expression such as \(\frac{2x}{x^2+1}\), consider \(t=x^2+1\).
  • Expression inside a square root: For an expression such as \(\sqrt{1+x^2}\), consider whether \(t=1+x^2\) simplifies the integral.
  • Trigonometric expressions: When an expression and its derivative appear together, a substitution involving that expression may simplify the integral.
Important observation: The derivative of the selected expression need not appear exactly as written. A non-zero constant multiple can usually be adjusted outside the integral.
🗒️ General Procedure
  1. Examine the integrand and identify a composite expression that may simplify the integral.
  2. Let that expression be a new variable, for example, \(t=g(x)\).
  3. Differentiate the substitution to find \(dt=g'(x)\,dx\).
  4. Express the required factors of the original integrand in terms of \(t\) and \(dt\).
  5. Rewrite the entire integral using the new variable.
  6. Evaluate the resulting integral using a standard formula.
  7. Substitute the original expression back in place of \(t\).
  8. Add the constant of integration \(C\) and, when appropriate, verify the answer by differentiation.
✏️ Example
1
Question
Evaluate
\[\int 2x(x^2+3)^4\,dx\]
The expression \(x^2+3\) is raised to a power, and its derivative, \(2x\), is present as a factor. This makes substitution the natural choice.
  1. 1

    Solution roadmap: Substitute \(t=x^2+3\), convert \(2x\,dx\) into \(dt\), apply the power rule and return to the original variable.

  1. Choose the substitution
    \[t=x^2+3\]
  2. Differentiate
    \[\frac{dt}{dx}=2x\]
  3. Therefore,
    \[dt=2x\,dx\]
  4. Rewrite the integral
    \[ \begin{aligned} I&=\int 2x(x^2+3)^4\,dx\\ &=\int t^4\,dt \end{aligned} \]
  5. Apply the power rule.
    \[ \begin{aligned} I&=\frac{t^5}{5}+C \end{aligned} \]
  6. Substitute back
    \boxed{I=\frac{(x^2+3)^5}{5}+C}
  7. <strong>Verification:</strong> Differentiating the result give
    \[ \begin{aligned} \frac{d}{dx}\left[\frac{(x^2+3)^5}{5}\right] &=\frac{1}{5}\cdot5(x^2+3)^4\cdot2x\\ &=2x(x^2+3)^4 \end{aligned} \]
    📝 The derivative matches the original integrand, so the result is correct.
2
Question
Evaluate
\[\int \frac{2x}{x^2+5}\,dx\]
The numerator is the derivative of the expression in the denominator. The logarithmic integration formula can therefore be applied after substitution.
  1. Substitute
    \[t=x^2+5\]
  2. Differentiating,
    \[dt=2x\,dx\]
  3. Transform the integral
    \[ \begin{aligned} I&=\int\frac{2x}{x^2+5}\,dx\\ &=\int\frac{1}{t}\,dt \end{aligned} \]
  4. Integrate and substitute back
    \[ \begin{aligned} I&=\ln|t|+C\\ &=\ln|x^2+5|+C \end{aligned} \]
  5. Since \(x^2+5>0\) for every real \(x\), the absolute value can be omitted.
    \[\boxed{I=\ln(x^2+5)+C}\]
  6. Verification:
    \[\frac{d}{dx}\ln(x^2+5)=\frac{2x}{x^2+5}\]
    📝 This confirms the result.
3
Question
Evaluate
\[\int \cos(3x+1)\,dx\]
The derivative of the angle \(3x+1\) is the constant \(3\). A constant adjustment is needed after substitution.
  1. Let
    \[t=3x+1\]
  2. Then
    \[dt=3\,dx\quad\Rightarrow\quad dx=\frac{dt}{3}\]
  3. Substitute into the integral
    \[ \begin{aligned} I&=\int\cos(3x+1)\,dx\\ &=\frac{1}{3}\int\cos t\,dt \end{aligned} \]
  4. Integrate and substitute back
    \[ \begin{aligned} I&=\frac{1}{3}\sin t+C\\ &=\frac{1}{3}\sin(3x+1)+C \end{aligned} \]
    \[ \boxed{I=\frac{1}{3}\sin(3x+1)+C} \]
  5. Verification:
    \[ \frac{d}{dx}\left[\frac{1}{3}\sin(3x+1)\right] =\cos(3x+1) \]
    📝 The result is correct
4
Question
Evaluate
\[\int x\sqrt{x^2+1}\,dx\]
The expression under the square root is \(x^2+1\), whose derivative is \(2x\). The integrand contains \(x\,dx\), which differs from the required differential by a factor of \(2\).
  1. Choose the substitution
    \[t=x^2+1\]
  2. Thus,
    \[dt=2x\,dx\quad\Rightarrow\quad x\,dx=\frac{dt}{2}\]
  3. Rewrite the integral
    \[ \begin{aligned} I&=\int x\sqrt{x^2+1}\,dx\\ &=\frac{1}{2}\int t^{1/2}\,dt \end{aligned} \]
  4. Apply the power rule
    \[ \begin{aligned} I&=\frac{1}{2}\cdot\frac{t^{3/2}}{3/2}+C\\ &=\frac{1}{3}t^{3/2}+C \end{aligned} \]
  5. Substitute back
    \[\boxed{I=\frac{(x^2+1)^{3/2}}{3}+C}\]
  6. Verification:
    \[ \begin{aligned} \frac{d}{dx}\left[\frac{(x^2+1)^{3/2}}{3}\right] &=\frac{1}{3}\cdot\frac{3}{2}(x^2+1)^{1/2}\cdot2x\\ &=x\sqrt{x^2+1} \end{aligned} \]
    📝 Hence, the answer is verified.
5
Question
Evaluate
\[\int\frac{1}{2x+7}\,dx\]
  1. Let
    \[t=2x+7\]
  2. Then
    \[ dt=2\,dx \qquad\Rightarrow\quad dx=\frac{dt}{2} \]
  3. Substitute
    \[ \begin{aligned} I&=\int\frac{1}{2x+7}\,dx\\ &=\frac{1}{2}\int\frac{1}{t}\,dt \end{aligned} \]
  4. Integrate and substitute back
    \[ \begin{aligned} I&=\frac{1}{2}\ln|t|+C\\ &=\frac{1}{2}\ln|2x+7|+C \end{aligned} \]
    \[ \boxed{I=\frac{1}{2}\ln|2x+7|+C} \]
    📝 The original integrand is undefined at \(x=-\frac{7}{2}\). The antiderivative is understood on any interval that does not cross this point.
🌟 Important Standard Forms Based on Substitution
❌ Common Mistakes
  • Forgetting to change the differential: If \(t=g(x)\), then \(dt=g'(x)\,dx\). The differential must be converted along with the expression.
  • Missing a constant factor: If \(t=3x+1\), then \(dx=\frac{dt}{3}\), not \(dt\).
  • Leaving the answer in two variables: For an indefinite integral originally expressed in \(x\), substitute back to express the final result in \(x\).
  • Applying the power rule incorrectly: The formula \(\int t^n\,dt=\frac{t^{n+1}}{n+1}+C\) is valid for \(n\ne-1\). When \(n=-1\), use \(\int\frac{1}{t}\,dt=\ln|t|+C\).
  • Ignoring the domain: Logarithmic integrals require a non-zero argument, and the antiderivative is considered on intervals where the original integrand is defined.
  • Choosing an unhelpful substitution: A substitution should simplify the integral, not merely replace one expression with another.
  • Omitting the constant of integration: Every indefinite integral must include \(C\).
⚡ Exam Tip
Exam Tips for CBSE and Competitive Examinations
🗒️ Practice Questions

Try the following questions using integration by substitution before checking the answers.

  1. \(\int 4x(x^2+2)^3\,dx\)
  2. \(\int \frac{3x^2}{x^3+4}\,dx\)
  3. \(\int e^{5x}\,dx\)
  4. \(\int \sin(4x-2)\,dx\)
  5. \(\int \frac{x}{\sqrt{x^2+9}}\,dx\)
Show answers
  1. \(\frac{(x^2+2)^4}{2}+C\)
  2. \(\ln|x^3+4|+C\)
  3. \(\frac{e^{5x}}{5}+C\)
  4. \(-\frac{1}{4}\cos(4x-2)+C\)
  5. \(\sqrt{x^2+9}+C\)
🔑 Key Takeaway
🗒️ Frequently Asked Questions (FAQs)
1. What is integration by substitution?

It is a method of integration in which a suitable expression is replaced by a new variable to simplify the integral. It is based on the chain rule of differentiation.

2. When should substitution be used?

Use substitution when the integrand contains a composite function and the derivative of its inner expression is also present, possibly multiplied by a constant.

3. Why is the differential changed during substitution?

The differential changes because the variables are related. If \(t=g(x)\), then \(dt=g'(x)\,dx\). This relationship ensures that the transformed integral is mathematically equivalent to the original one.

4. Must the final answer be expressed in the original variable?

For an indefinite integral originally written in \(x\), the final answer should normally be expressed in \(x\), so the substitution must be reversed before presenting the result.

5. What is the difference between substitution and integration by parts?

Substitution simplifies an integral through a change of variable and reverses the chain rule. Integration by parts is based on the product rule and is particularly useful for products of functions.

6. Is substitution always the best method?

No. Some integrals require algebraic simplification, partial fractions, integration by parts or trigonometric identities. The appropriate method depends on the structure of the integrand.

7. How can an answer obtained by substitution be verified?

Differentiate the proposed antiderivative. If its derivative equals the original integrand on the relevant domain, the answer is verified.

∫

Example 4

📋
Table of Contents
2 sections
❓ Question
Integrate the following functions with respect to \(x\), using the method of substitution.
  1. \(\sin mx\)
  2. \(2x\sin(x^2+1)\)
  3. \(\dfrac{\tan^4\sqrt{x}\,\sec^2\sqrt{x}}{\sqrt{x}}\)
  4. \(\dfrac{\sin(\tan^{-1}x)}{1+x^2}\)
🧩 Solution
Part (i)
Integrate \(\sin mx\)
  1. 💡
    Concept
    The argument of the sine function is \(mx\), where \(m\) is a constant. Since the derivative of \(mx\) is \(m\), substitution converts the integral into the standard form \(\int\sin t\,dt\). The factor \(\frac{1}{m}\) must be included when \(m\ne0\).
    🗺️ Roadmap
    1

    Substitute \(t=mx\).

    2

    Differentiate to obtain \(dx=\frac{dt}{m}\).

    3

    Evaluate the resulting trigonometric integral.

    4

    Replace \(t\) with \(mx\) in the final expression.

  2. Choose the substitution
    \[t=mx\]
  3. Differentiating both sides with respect to \(x\), we obtain
    \frac{dt}{dx}=m
  4. Therefore,
    \[dt=m\,dx\]
    dx=\frac{1}{m}\,dt,\qquad m\ne0
  5. Substitute into the integral
    \[ \begin{aligned} I&=\int\sin mx\,dx\\ &=\int\sin t\left(\frac{1}{m}\,dt\right)\\ &=\frac{1}{m}\int\sin t\,dt \end{aligned} \]
  6. Apply the standard formula
    \[\int\sin t\,dt=-\cos t+C\]
  7. Hence,
    \[I=-\frac{1}{m}\cos t+C\]
  8. Substitute \(t=mx\).
    \[\boxed{\int\sin mx\,dx=-\frac{\cos mx}{m}+C,\quad m\ne0}\]
  9. Verification by Differentiating the obtained result
    \[ \begin{aligned} \frac{d}{dx}\left(-\frac{\cos mx}{m}\right) &=-\frac{1}{m}(-\sin mx)\cdot m\\ &=\sin mx \end{aligned} \]
    📝 The derivative equals the original integrand. Therefore, the answer is correct.

    Special case: If \(m=0\), the integrand is \(\sin 0=0\), so the integral is simply \(C\).
  10. ⭐ Exam tip: For a linear argument \(ax+b\), remember the general result
    \[\boxed{\int\sin(ax+b)\,dx=-\frac{\cos(ax+b)}{a}+C,\quad a\ne0}\]
Part (ii)
\(2x\sin(x^2+1)\)
  1. 💡
    Concept
    The expression \(x^2+1\) occurs inside the sine function. Its derivative is \(2x\), which appears as a factor in the integrand. This is a direct indication that substitution will simplify the integral.
    🗺️ Roadmap
    1

    Let \(t=x^2+1\).

    2

    Use \(dt=2x\,dx\).

    3

    Integrate \(\sin t\).

    4

    Substitute the original expression back.

  2. Choose the substitution.
    \[t=x^2+1\]
  3. Differentiating with respect to \(x\),
    \[\frac{dt}{dx}=2x\]
  4. Thus,
    \[dt=2x\,dx\]
    📝 Notice that the complete factor \(2x\,dx\) is present in the original integral. There is no need to divide by \(2x\).
  5. Transform the integral.
    \[ \begin{aligned} I&=\int 2x\sin(x^2+1)\,dx\\ &=\int\sin t\,dt \end{aligned} \]
  6. Integrate
    \[I=-\cos t+C\]
  7. Substitute \(t=x^2+1\).
    \[\boxed{\int 2x\sin(x^2+1)\,dx=-\cos(x^2+1)+C}\]
  8. Verification
    \[ \begin{aligned} \frac{d}{dx}\left[-\cos(x^2+1)\right] &=\sin(x^2+1)\cdot2x\\ &=2x\sin(x^2+1) \end{aligned} \]
    📝 The result is verified.
  9. ⭐ Important observation: The differential relation is \(dt=2x\,dx\). Writing \(dx=\frac{dt}{2x}\) is unnecessary here and can introduce avoidable algebraic complications. Always use the form that matches the factors already present in the integrand.
Part (iii)
\(\dfrac{\tan^4\sqrt{x}\,\sec^2\sqrt{x}}{\sqrt{x}}\)
  1. 💡
    Concept

    This integral requires two successive substitutions. First, the square root \(\sqrt{x}\) is replaced with a new variable to simplify the argument of the trigonometric functions. Next, the tangent of that variable is substituted because the derivative of \(\tan t\) is \(\sec^2t\).

    This is an example of successive substitution, in which one change of variable prepares the integral for another.

    🗺️ Roadmap
    1

    Set \(t=\sqrt{x}\) and obtain \(dx=2t\,dt\).

    2

    Cancel the factor \(t\) against the denominator.

    3

    Set \(u=\tan t\), so that \(du=\sec^2t\,dt\).

    4

    Apply the power rule and reverse both substitutions.

  2. Make the first substitution
    \[t=\sqrt{x}\]
  3. Squaring both sides gives
    \[x=t^2\]
  4. Differentiating with respect to \(t\),
    \[\frac{dx}{dt}=2t\]
  5. Therefore,
    \[dx=2t\,dt\]
  6. Since \(t=\sqrt{x}\), this is equivalent to \(dx=2\sqrt{x}\,dt\). Substitute it into the original integral.
    \[ \begin{aligned} I &=\int\frac{\tan^4\sqrt{x}\,\sec^2\sqrt{x}}{\sqrt{x}}\,dx\\ &=\int\frac{\tan^4t\,\sec^2t}{t}(2t\,dt) \end{aligned} \]
  7. Cancel \(t\) from the numerator and denominator:
    \[I=2\int\tan^4t\,\sec^2t\,dt\]
  8. Make the second substitution.
    \[u=\tan t\]
  9. Differentiating,
    \[du=\sec^2t\,dt\]
  10. The integral now becomes
    \[ \begin{aligned} I&=2\int\tan^4t\,\sec^2t\,dt\\ &=2\int u^4\,du \end{aligned} \]
  11. Apply the power rule.
    \[ \begin{aligned} I&=2\left(\frac{u^5}{5}\right)+C\\ &=\frac{2}{5}u^5+C \end{aligned} \]
  12. Reverse the second substitution
    Since \(u=\tan t\),
    \[I=\frac{2}{5}\tan^5t+C\]
  13. Reverse the first substitution.
    Since \(t=\sqrt{x}\), the final result is
    \[ \boxed{ \int\frac{\tan^4\sqrt{x}\,\sec^2\sqrt{x}}{\sqrt{x}}\,dx =\frac{2}{5}\tan^5\sqrt{x}+C } \]
  14. Verification: Differentiate the result using the chain rule:
    \[ \begin{aligned} \frac{d}{dx}\left(\frac{2}{5}\tan^5\sqrt{x}\right) &=\frac{2}{5}\cdot5\tan^4\sqrt{x}\\ &\quad\cdot\sec^2\sqrt{x}\cdot\frac{1}{2\sqrt{x}}\\ &=\frac{\tan^4\sqrt{x}\,\sec^2\sqrt{x}}{\sqrt{x}} \end{aligned} \]
    📝 The derivative matches the original integrand.
  15. Domain note
    📝 The original integrand requires \(x>0\), and \(\tan\sqrt{x}\) and \(\sec\sqrt{x}\) must be defined. The antiderivative is understood on intervals that avoid the singularities of these trigonometric functions.
Part (iv)
\(\dfrac{\sin(\tan^{-1}x)}{1+x^2}\)
  1. 💡
    Concept
    The expression \(\tan^{-1}x\) is an inverse trigonometric function. Its derivative is
    \[\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}\]
    This derivative appears exactly in the denominator of the integrand. Therefore, substituting \(t=\tan^{-1}x\) transforms the integral into \(\int\sin t\,dt\).
    🗺️ Roadmap
    1

    Let \(t=\tan^{-1}x\).

    2

    Differentiate to obtain \(dt=\frac{dx}{1+x^2}\).

    3

    Rewrite the integral in terms of \(t\).

    4

    Integrate and substitute back.

  2. Choose the substitution
    \[t=\tan^{-1}x\]
  3. Differentiating both sides with respect to \(x\),
    \[\frac{dt}{dx}=\frac{1}{1+x^2}\]
  4. Therefore,
    \[dt=\frac{dx}{1+x^2}\]
  5. Equivalently,
    \[dx=(1+x^2)\,dt\]
  6. Substitute into the integral
    \[ \begin{aligned} I&=\int\frac{\sin(\tan^{-1}x)}{1+x^2}\,dx\\ &=\int\sin t\,\frac{dx}{1+x^2}\\ &=\int\sin t\,dt \end{aligned} \]
  7. Integrate
    \[I=-\cos t+C\]
  8. Substitute \(t=\tan^{-1}x\)
    \[ \boxed{ I=-\cos(\tan^{-1}x)+C } \]
  9. The result can also be expressed without an inverse trigonometric function. Let \(\theta=\tan^{-1}x\). Then \(\tan\theta=x\), and \(\theta\) lies in \((-\frac{\pi}{2},\frac{\pi}{2})\), where \(\cos\theta>0\).
    Using a right-triangle interpretation, take the opposite side to be \(x\), the adjacent side to be \(1\), and the hypotenuse to be \(\sqrt{1+x^2}\). Hence,
    \[\cos(\tan^{-1}x)=\frac{1}{\sqrt{1+x^2}}\]
  10. Therefore, an equivalent form of the answer is
    \[\boxed{I=-\frac{1}{\sqrt{1+x^2}}+C}\]
  11. Verification: Differentiate the simplified expression:
    \[ \begin{aligned} \frac{d}{dx}\left[-(1+x^2)^{-1/2}\right] &=-\left(-\frac{1}{2}\right)(1+x^2)^{-3/2}(2x)\\ &=\frac{x}{(1+x^2)^{3/2}} \end{aligned} \]
  12. To verify that this equals the original integrand, use
    \[\sin(\tan^{-1}x)=\frac{x}{\sqrt{1+x^2}}\]
  13. Thus
    \[ \frac{\sin(\tan^{-1}x)}{1+x^2} =\frac{x}{(1+x^2)^{3/2}} \]
    📝 The derivative agrees with the original integrand, confirming the result.
∫

Important Integrals Involving Trigonometric Functions

📋
Table of Contents
6 sections
🗺️ Overview

Several trigonometric integrals can be evaluated conveniently by using the method of substitution. These standard results are important in Class 12 Mathematics because they simplify the evaluation of more complicated integrals and are frequently used in subsequent exercises without repeating their derivations.

In this section, we derive four important standard integrals: the integrals of \(\tan x\), \(\cot x\), \(\sec x\) and \(\operatorname{cosec} x\). Each derivation demonstrates how a suitable algebraic rearrangement or substitution transforms the integrand into a standard logarithmic form.

🗂️ Important Integrals Involving Trigonometric Functions
  • (I) INTEGRAL OF \(\tan x\)
    🎯To Prove
    \[\boxed{\int\tan x\,dx=\ln|\sec x|+C}\]
    ⭐HighlightUsing the trigonometric identity \(\tan x=\frac{\sin x}{\cos x}\), the integral can be rewritten as a quotient. Since the derivative of \(\cos x\) is \(-\sin x\), the substitution \(t=\cos x\) converts the integral into the standard logarithmic form \(\int\frac{1}{t}\,dt\).
    🧮ProofStarting with the given integral,
    \[I=\int\tan x\,dx\]
    Using \(\tan x=\frac{\sin x}{\cos x}\), we obtain
    \[I=\int\frac{\sin x}{\cos x}\,dx\]
    Step 1: Make the substitution.
    \[t=\cos x\]
    Differentiating both sides,
    \[dt=-\sin x\,dx\]
    Therefore,
    \[\sin x\,dx=-dt\]
    Step 2: Substitute into the integral.
    \[ \begin{aligned} I&=\int\frac{\sin x}{\cos x}\,dx\\ &=\int\frac{-dt}{t}\\ &=-\int\frac{1}{t}\,dt \end{aligned} \]
    Step 3: Apply the logarithmic integral formula.
    \[\int\frac{1}{t}\,dt=\ln|t|+C\]
    Hence,
    \[I=-\ln|t|+C\]
    Substituting \(t=\cos x\),
    \[I=-\ln|\cos x|+C\]
    Using the logarithmic identity \(-\ln|a|=\ln\left|\frac{1}{a}\right|\), we obtain
    \[ \begin{aligned} I&=\ln\left|\frac{1}{\cos x}\right|+C\\ &=\ln|\sec x|+C \end{aligned} \]
    Therefore,
    \[\boxed{\int\tan x\,dx=\ln|\sec x|+C}\]
    🏁Hence Proved
  • (II) INTEGRAL OF \(\cot x\)
    🎯To Prove
    \[\boxed{\int\cot x\,dx=\ln|\sin x|+C}\]
    ⭐HighlightExpress \(\cot x\) as \(\frac{\cos x}{\sin x}\). The numerator is the derivative of the denominator, making the substitution \(t=\sin x\) suitable.
    🧮ProofWe have
    \[I=\int\cot x\,dx\]
    Using the identity \(\cot x=\frac{\cos x}{\sin x}\),
    \[I=\int\frac{\cos x}{\sin x}\,dx\]
    Step 1: Choose the substitution.
    \[t=\sin x\]
    Differentiating,
    \[dt=\cos x\,dx\]
    Step 2: Transform the integral.
    \[ \begin{aligned} I&=\int\frac{\cos x}{\sin x}\,dx\\ &=\int\frac{1}{t}\,dt \end{aligned} \]
    Step 3: Integrate
    \[I=\ln|t|+C\]
    Substituting \(t=\sin x\), we obtain
    \[\boxed{\int\cot x\,dx=\ln|\sin x|+C}\]
    🏁Hence Proved
  • (III) INTEGRAL OF \(\sec x\)
    🎯To Prove
    \[\boxed{\int\sec x\,dx=\ln|\sec x+\tan x|+C}\]
    ⭐HighlightThe integral of \(\sec x\) is not obtained as directly as those of \(\tan x\) and \(\cot x\). A useful technique is to multiply and divide the integrand by \(\sec x+\tan x\). The derivative of this expression is proportional to the same expression, allowing the integral to be reduced to a logarithmic form.
    🧮ProofConsider
    \[I=\int\sec x\,dx\]
    Step 1: Multiply and divide by \(\sec x+\tan x\).
    \[I=\int\sec x\,\frac{\sec x+\tan x}{\sec x+\tan x}\,dx\]
    Rearranging,
    \[I=\int\frac{\sec^2x+\sec x\tan x}{\sec x+\tan x}\,dx\]
    Step 2: Identify the derivative in the numerator. Recall the standard derivatives
    \[\frac{d}{dx}(\sec x)=\sec x\tan x\]
    \[\frac{d}{dx}(\tan x)=\sec^2x\]
    Therefore,
    \[\frac{d}{dx}(\sec x+\tan x)=\sec x\tan x+\sec^2x\]
    This is precisely the numerator of the transformed integral

    Step 3: Apply substitution.
    \[t=\sec x+\tan x\]
    Then
    \[dt=(\sec x\tan x+\sec^2x)\,dx\]
    The integral becomes
    \[I=\int\frac{1}{t}\,dt\]
    Step 4: Integrate and substitute back.
    \[I=\ln|t|+C\]
    Hence,
    \[ \boxed{\int\sec x\,dx=\ln|\sec x+\tan x|+C}\]
    🏁Hence Proved
  • (IV) INTEGRAL OF \(\operatorname{cosec} x\)
    🎯To Prove
    \[\boxed{\int\operatorname{cosec} x\,dx=\ln|\operatorname{cosec} x-\cot x|+C}\]
    ⭐HighlightSimilar to the integral of \(\sec x\), we multiply and divide by a carefully chosen expression. Here, the useful expression is \(\operatorname{cosec} x-\cot x\), because its derivative is proportional to itself.
    🧮ProofConsider
    \[I=\int\operatorname{cosec} x\,dx\]
    Step 1: Multiply and divide by \(\operatorname{cosec} x-\cot x\).
    \[I=\int\operatorname{cosec} x\,\frac{\operatorname{cosec} x-\cot x}{\operatorname{cosec} x-\cot x}\,dx\]
    Rearranging the numerator,
    \[I=\int\frac{\operatorname{cosec}^2x-\operatorname{cosec} x\cot x}{\operatorname{cosec} x-\cot x}\,dx\]
    Step 2: Identify the derivative in the numerator.

    Using the standard derivatives,
    \[\frac{d}{dx}(\operatorname{cosec} x)=-\operatorname{cosec} x\cot x\]
    \[\frac{d}{dx}(-\cot x)=\operatorname{cosec}^2x\]
    Therefore,
    \[ \frac{d}{dx}(\operatorname{cosec} x-\cot x) =-\operatorname{cosec} x\cot x+\operatorname{cosec}^2x \]
    This is the numerator of the transformed integral. Step 3: Apply substitution.
    \[t=\operatorname{cosec} x-\cot x\]
    Then
    \[dt=(\operatorname{cosec}^2x-\operatorname{cosec} x\cot x)\,dx\]
    Consequently,
    \[I=\int\frac{1}{t}\,dt\]
    Step 4: Integrate and substitute back.
    \[I=\ln|t|+C\]
    Thus,
    \[\boxed{\int\operatorname{cosec} x\,dx=\ln|\operatorname{cosec} x-\cot x|+C}\]
    🏁Hence Proved
📝 Summary of Important Trigonometric Integrals
⚡ Important Observations and Exam Tips
🔑 Key Takeaway
🗒️ Frequently Asked Questions (FAQs)

1. What is the integral of \(\tan x\)?

The integral is \(\int\tan x\,dx=\ln|\sec x|+C\), equivalently \(-\ln|\cos x|+C\).

2. What is the integral of \(\cot x\)?

The integral is \(\int\cot x\,dx=\ln|\sin x|+C\).

3. How is the integral of \(\sec x\) derived?

Multiply and divide by \(\sec x+\tan x\). The numerator then becomes the derivative of \(\sec x+\tan x\), producing a logarithmic integral.

4. How is the integral of \(\operatorname{cosec} x\) derived?

Multiply and divide by \(\operatorname{cosec} x-\cot x\). Its derivative appears in the numerator, reducing the integral to \(\int\frac{1}{u}\,du\).

5. Why is the absolute value used in these formulas?

Because the real logarithm \(\ln|u|\) is defined for any non-zero real \(u\). The absolute value makes the antiderivative formula valid on intervals where \(u\) is non-zero.

∫

Example 5

📋
Table of Contents
6 sections
❓ Question
Find the following integrals using suitable substitutions and trigonometric identities.
  1. \(\quad\int \sin^3x\cos^2x\,dx\)
  2. \(\quad\int\frac{\sin x}{\sin(x+a)}\,dx\)
  3. \(\quad \int\frac{1}{1+\tan x}\,dx\)
🧩 Solution
Part (i)
\(\int \sin^3x\cos^2x\,dx\)
  1. 💡
    Concept
    The integrand contains an odd power of \(\sin x\). Separate one factor of \(\sin x\), convert the remaining even power using \(\sin^2x=1-\cos^2x\), and substitute \(t=\cos x\). This converts the trigonometric integral into a polynomial integral.
    🗺️ Roadmap
    1

    Write \(\sin^3x=\sin^2x\sin x\).

    2

    Use \(\sin^2x=1-\cos^2x\).

    3

    Substitute \(t=\cos x\), so \(dt=-\sin x\,dx\).

    4

    Expand, integrate using the power rule, and substitute back.

  2. Step 1: Rewrite the integral.
    \[ \begin{aligned} I&=\int\sin^3x\cos^2x\,dx\\ &=\int\sin^2x\cos^2x\sin x\,dx \end{aligned} \]
  3. Using the identity
    \[\sin^2x=1-\cos^2x\]
  4. we obtain
    \[I=\int(1-\cos^2x)\cos^2x\sin x\,dx\]
  5. Step 2: Make the substitution.
    \[t=\cos x\]
  6. Differentiating,
    \[dt=-\sin x\,dx\]
  7. Therefore,
    \[\sin x\,dx=-dt\]
  8. Step 3: Transform the integral.
    \[ \begin{aligned} I&=\int(1-\cos^2x)\cos^2x\sin x\,dx\\ &=-\int(1-t^2)t^2\,dt \end{aligned} \]
  9. Step 4: Expand the polynomial.
    \[ \begin{aligned} I&=-\int(t^2-t^4)\,dt\\ &=-\int t^2\,dt+\int t^4\,dt \end{aligned} \]
  10. Step 5: Integrate using the power rule.
    \[ \begin{aligned} I&=-\frac{t^3}{3}+\frac{t^5}{5}+C\\ &=\frac{t^5}{5}-\frac{t^3}{3}+C \end{aligned} \]
  11. Step 6: Substitute \(t=\cos x\).
    \[ \boxed{ \int\sin^3x\cos^2x\,dx =\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C } \]
  12. Verification: Differentiate the result
    \[ \begin{aligned} \frac{d}{dx}\left( \frac{\cos^5x}{5}-\frac{\cos^3x}{3} \right) &=-\cos^4x\sin x+\cos^2x\sin x\\ &=\cos^2x\sin x(1-\cos^2x)\\ &=\cos^2x\sin^3x \end{aligned} \]
    📝 This equals the original integrand, confirming the answer.
  13. ⭐ Exam tip: When one trigonometric function has an odd power, save one factor of that function for the differential and convert the remaining even power using a Pythagorean identity. For example, an odd power of \(\sin x\) often suggests \(t=\cos x\), while an odd power of \(\cos x\) often suggests \(t=\sin x\).]
Part (ii)
\(\int\frac{\sin x}{\sin(x+a)}\,dx\)
  1. 💡
    Concept
    The denominator contains the shifted angle \(x+a\). Substituting \(t=x+a\) makes the denominator \(\sin t\). The numerator becomes \(\sin(t-a)\), which can be expanded using the sine difference identity. The resulting integral separates into a constant term and a standard cotangent-related logarithmic integral.
    Here, \(a\) is a constant
    🗺️ Roadmap
    1

    Set \(t=x+a\), so \(x=t-a\) and \(dx=dt\).

    2

    Expand \(\sin(t-a)\) using the sine difference formula.

    3

    Divide each term by \(\sin t\).

    4

    Integrate the constant and cotangent terms separately.

    5

    Substitute \(t=x+a\) back into the answer.

  2. Choose the substitution.
    \[t=x+a\]
  3. Since \(a\) is constant,
    dt=dx
  4. Also,
    \[x=t-a\]
  5. Therefore,
    \[I=\int\frac{\sin(t-a)}{\sin t}\,dt\]
  6. Apply the sine difference identity.
    \[\sin(A-B)=\sin A\cos B-\cos A\sin B\]
  7. Thus,
    \[\sin(t-a)=\sin t\cos a-\cos t\sin a\]
  8. Substituting this expression gives
    \[ \begin{aligned} I&=\int\frac{\sin t\cos a-\cos t\sin a}{\sin t}\,dt\\ &=\int\left(\cos a-\sin a\frac{\cos t}{\sin t}\right)\,dt\\ &=\int(\cos a-\sin a\cot t)\,dt \end{aligned} \]
  9. Separate the integrals.
    \[I=\cos a\int dt-\sin a\int\cot t\,dt\]
    📝 Since \(a\) is constant, \(\cos a\) and \(\sin a\) can be taken outside the integrals.
  10. Apply the standard integral formulas
    \[\int dt=t+C\]
    and
    \[\int\cot t\,dt=\ln|\sin t|+C\]
  11. Therefore,
    \[ \begin{aligned} I&=t\cos a-\sin a\ln|\sin t|+C \end{aligned} \]
  12. Substitute \(t=x+a\)
    \[I=(x+a)\cos a-\sin a\ln|\sin(x+a)|+C\]
  13. Expanding the first term gives
    \[I=x\cos a+a\cos a-\sin a\ln|\sin(x+a)|+C\]
  14. Since \(a\cos a\) is constant with respect to \(x\), it can be absorbed into the arbitrary constant of integration.<br><br> Hence, the final answer is
    \[ \boxed{ \int\frac{\sin x}{\sin(x+a)}\,dx =x\cos a-\sin a\ln|\sin(x+a)|+C } \]
  15. Verification: Differentiate the result
    \[ \begin{aligned} \frac{d}{dx}\left[ x\cos a-\sin a\ln|\sin(x+a)| \right] &=\cos a-\sin a\cot(x+a)\\ &=\frac{\sin(x+a)\cos a-\cos(x+a)\sin a} {\sin(x+a)}\\ &=\frac{\sin((x+a)-a)}{\sin(x+a)}\\ &=\frac{\sin x}{\sin(x+a)} \end{aligned} \]
    📝 The derivative equals the original integrand.
  16. Domain note
    The original integrand is undefined where \(\sin(x+a)=0\). The antiderivative applies on intervals that do not cross these points.
Part (iii)
\(\int\frac{1}{1+\tan x}\,dx\)
  1. 💡
    Concept
    First express \(\tan x\) as \(\frac{\sin x}{\cos x}\). Then combine the terms into a single fraction. By splitting the numerator into two suitable parts, one integral becomes a simple constant multiple of \(x\), while the other takes the logarithmic form \(\int\frac{f'(x)}{f(x)}\,dx\).
    🗺️ Roadmap
    1

    Rewrite the integrand using sine and cosine.

    2

    Split the numerator into \((\sin x+\cos x)+(\cos x-\sin x)\).

    3

    Separate the integral into two parts.

    4

    Use \(t=\sin x+\cos x\) for the second integral.

    5

    Combine the results and state the domain restrictions.

  2. Rewrite the integrand
    \[ \begin{aligned} I&=\int\frac{1}{1+\tan x}\,dx\\ &=\int\frac{1}{1+\frac{\sin x}{\cos x}}\,dx\\ &=\int\frac{\cos x}{\sin x+\cos x}\,dx \end{aligned} \]
    📝 This expression is considered where the original integrand is defined.
  3. Split the numerator suitably. Write
    \[\cos x=\frac{1}{2}\left[(\sin x+\cos x)+(\cos x-\sin x)\right]\]
  4. Substituting this into the integral,
    \[ \begin{aligned} I&=\frac{1}{2}\int \frac{(\sin x+\cos x)+(\cos x-\sin x)} {\sin x+\cos x}\,dx\\ &=\frac{1}{2}\int dx+ \frac{1}{2}\int \frac{\cos x-\sin x}{\sin x+\cos x}\,dx \end{aligned} \]
  5. Therefore,
    \[I=\frac{x}{2}+\frac{1}{2}\int\frac{\cos x-\sin x}{\sin x+\cos x}\,dx\]
  6. Use substitution for the remaining integral.
    \[t=\sin x+\cos x\]
  7. Differentiating,
    \[dt=(\cos x-\sin x)\,dx\]
  8. Thus,
    \[ \begin{aligned} \frac{1}{2}\int \frac{\cos x-\sin x}{\sin x+\cos x}\,dx &=\frac{1}{2}\int\frac{1}{t}\,dt\\ &=\frac{1}{2}\ln|t|+C \end{aligned} \]
  9. Substitute back.
    \[ \frac{1}{2}\int \frac{\cos x-\sin x}{\sin x+\cos x}\,dx =\frac{1}{2}\ln|\sin x+\cos x|+C \]
  10. Combining both parts gives
    \[ \boxed{ \int\frac{1}{1+\tan x}\,dx =\frac{x}{2}+\frac{1}{2}\ln|\sin x+\cos x|+C } \]
  11. Verification: Differentiate the result
    \[ \begin{aligned} \frac{d}{dx}\left[ \frac{x}{2}+\frac{1}{2}\ln|\sin x+\cos x| \right] &=\frac{1}{2}+ \frac{1}{2}\frac{\cos x-\sin x}{\sin x+\cos x}\\ &=\frac{1}{2} \left[ \frac{\sin x+\cos x+\cos x-\sin x} {\sin x+\cos x} \right]\\ &=\frac{\cos x}{\sin x+\cos x}\\ &=\frac{1}{1+\tan x} \end{aligned} \]
    📝 Hence, the result is correct
  12. Domain note
    The original integrand requires \(\cos x\ne0\) and \(1+\tan x\ne0\). Equivalently, \(\cos x\ne0\) and \(\sin x+\cos x\ne0\). The antiderivative is valid on intervals avoiding these excluded points.
⚖️ Comparison of the Three Techniques
Part Main technique Key substitution or identity
(i) Trigonometric identity followed by substitution \(t=\cos x\), using \(\sin^2x=1-\cos^2x\)
(ii) Shifted-angle substitution and angle-difference identity \(t=x+a\), using \(\sin(t-a)\)
(iii) Algebraic splitting followed by logarithmic substitution \(t=\sin x+\cos x\)
❌ Common Mistakes
  • Forgetting the negative sign when using \(t=\cos x\), because \(dt=-\sin x\,dx\).
  • Confusing \(\cot t\) with \(\cot^{-1}t\). They represent different functions.
  • Forgetting that constants such as \(a\cos a\) can be absorbed into the arbitrary constant \(C\).
  • Splitting the numerator incorrectly. Verify that the proposed sum reproduces the original numerator.
  • Omitting the absolute value in logarithmic expressions when the argument can be negative.
  • Ignoring the points where the original integrand is undefined.
🌟 Significance

These integrals develop essential problem-solving skills for Class 12 Mathematics. The first tests the use of trigonometric identities with substitution, the second combines angle transformations with the standard integral of \(\cot x\), and the third tests algebraic manipulation and recognition of a logarithmic derivative.

In written examinations, show the identity used, the substitution, the differential conversion and the integration steps. In competitive examinations, recognising the structure of the integrand quickly can reduce lengthy calculations and help avoid sign errors.

🔑 Key Takeaway
∫

Integration Using Trigonometric Identities

📋
Table of Contents
1 sections
🗂️ Types / Category

When an integrand contains powers or combinations of trigonometric functions, direct application of standard integration formulas may not be possible. In such cases, we simplify the integrand using known trigonometric identities and then evaluate the resulting integral.

This method is called integration using trigonometric identities. It is particularly useful for integrating even powers of sine and cosine, products of trigonometric functions, and expressions involving multiple angles.

Important Trigonometric Identities Used in Integration

The following identities are frequently used to simplify integrands:

Power-reduction identities
\[ \sin^2x=\frac{1-\cos2x}{2} \]
\[ \cos^2x=\frac{1+\cos2x}{2} \]
Double-angle identity
\[\sin2x=2\sin x\cos x\]
Product-to-sum identities
\[ \sin A\cos B =\frac{1}{2}\left[\sin(A+B)+\sin(A-B)\right] \]
\[ \cos A\cos B =\frac{1}{2}\left[\cos(A+B)+\cos(A-B)\right] \]
\[ \sin A\sin B =\frac{1}{2}\left[\cos(A-B)-\cos(A+B)\right] \]
📌
Note
These identities transform complicated trigonometric expressions into forms that can be integrated using standard formulae.
∫

Example 6

📋
Table of Contents
8 sections
❓ Question
Evaluate
\[I=\int\cos^2x\,dx\]
💡 Concept
🗺️ Roadmap

  1. Apply the power-reduction identity for \(\cos^2x\).

  2. Separate the integral into two terms.

  3. Integrate the constant term.

  4. Use substitution to integrate the double-angle term.

  5. Combine the results and add the constant of integration.

🧩 Solution
Part (a)
  1. Write the given integral
    \[I=\int\cos^2x\,dx\]
  2. Apply the trigonometric identity
  3. ⭐ Using
    \[\cos2x=2\cos^2x-1\]
  4. we obtain
    \[\cos^2x=\frac{1+\cos2x}{2}\]
  5. Substituting into the integral,
    \[I=\int\frac{1+\cos2x}{2}\,dx\]
  6. Separating the terms,
    \[I=\frac{1}{2}\int dx+\frac{1}{2}\int\cos2x\,dx\]
  7. Integrate the first term.
    \frac{1}{2}\int dx=\frac{x}{2}
  8. Integrate the second term using substitution.
    Consider
    \[J=\frac{1}{2}\int\cos2x\,dx\]
  9. Let
    \[t=2x\]
  10. Differentiating both sides,
    \[dt=2\,dx\]
  11. Therefore,
    \[dx=\frac{dt}{2}\]
  12. Substituting,
    \[ \begin{aligned} J&=\frac{1}{2}\int\cos t\left(\frac{dt}{2}\right)\\ &=\frac{1}{4}\int\cos t\,dt\\ &=\frac{1}{4}\sin t+C \end{aligned} \]
  13. Since \(t=2x\),
    \[J=\frac{\sin2x}{4}\]
  14. Combine the results
    \[ \begin{aligned} I&=\frac{x}{2}+\frac{\sin2x}{4}+C \end{aligned} \]
  15. Hence, the final answer is
    \[\boxed{\int\cos^2x\,dx=\frac{x}{2}+\frac{\sin2x}{4}+C}\]
  16. Verification by Differentiation
    \[ \begin{aligned} \frac{d}{dx}\left(\frac{x}{2}+\frac{\sin2x}{4}\right) &=\frac{1}{2}+\frac{1}{4}(2\cos2x)\\ &=\frac{1}{2}+\frac{\cos2x}{2}\\ &=\frac{1+\cos2x}{2}\\ &=\cos^2x \end{aligned} \]
    📝 The derivative equals the original integrand, confirming that the answer is correct.
  17. Using the identity \(\sin2x=2\sin x\cos x\), the result can also be expressed as
    \[\boxed{\int\cos^2x\,dx=\frac{x}{2}+\frac{\sin x\cos x}{2}+C}\]
❌ Common Mistakes
  • Using the wrong identity: Remember that \(\cos^2x=\frac{1+\cos2x}{2}\), whereas \(\sin^2x=\frac{1-\cos2x}{2}\).
  • Missing the chain-rule factor: When integrating \(\cos2x\), the result is \(\frac{\sin2x}{2}+C\), not simply \(\sin2x+C\).
  • Incorrect sign: The integral of \(\cos2x\) is positive because the derivative of \(\sin2x\) is \(2\cos2x\).
  • Forgetting the constant: Include \(C\) in the final indefinite integral.
⚡ Exam Tip
❓ Practice Questions

Apply suitable trigonometric identities to evaluate the following integrals:

  1. \(\int\sin^2x\,dx\)
  2. \(\int\cos^2 3x\,dx\)
  3. \(\int\sin x\cos x\,dx\)
Show answers
  1. \(\frac{x}{2}-\frac{\sin2x}{4}+C\)
  2. \(\frac{x}{2}+\frac{\sin6x}{12}+C\)
  3. \(\frac{\sin^2x}{2}+C\)
🔑 Key Takeaway
∫

Integrals of Some Particular Functions

📋
Table of Contents
5 sections
🗺️ Overview

Several integrals involving rational functions, quadratic expressions and square roots occur repeatedly in Class 12 Mathematics. These integrals can often be evaluated using factorisation, partial fractions, completing the square, trigonometric substitution and the method of substitution.

The following standard results are especially important for solving NCERT exercises and more complicated integration problems. Once their derivations are understood, they can be used directly without repeating the entire proof each time.

Important Notation and Conditions

  • Unless otherwise stated, \(a\ne0\) in formulas involving division by \(a\).
  • Logarithmic antiderivatives are written with absolute values where necessary.
  • Each formula applies on intervals where the original integrand is defined.
  • The constant of integration is denoted by \(C\).
🗂️ Types / Category
  • (I) INTEGRAL OF \(\displaystyle\frac{1}{x^2-a^2}\)
    ⭐HighlightStandard formula
    \[\boxed{\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C}\]
    where \(a\ne0\) and \(x\ne\pm a\).
    💡ConceptThe denominator is a difference of two squares. Factorising it produces two linear factors, allowing the integrand to be expressed as a difference of simpler fractions. This is an application of partial fractions.
    🧩Solutionlet
    \[I=\int\frac{dx}{x^2-a^2}\]
    Step 1: Factorise the denominator.
    \[x^2-a^2=(x-a)(x+a)\]
    Therefore,
    \[I=\int\frac{dx}{(x-a)(x+a)}\]
    Step 2: Split the fraction.

    Observe that
    \[ \frac{1}{x^2-a^2} =\frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right) \]
    This follows because
    \[ \frac{1}{x-a}-\frac{1}{x+a} =\frac{2a}{(x-a)(x+a)} \]
    Hence,
    \[ I=\frac{1}{2a} \int\left(\frac{1}{x-a}-\frac{1}{x+a}\right)dx \]
    Step 3: Integrate each term.
    \[ \begin{aligned} I&=\frac{1}{2a} \left[\ln|x-a|-\ln|x+a|\right]+C\\ &=\frac{1}{2a} \ln\left|\frac{x-a}{x+a}\right|+C \end{aligned} \]
    Therefore,
    \[ \boxed{ \int\frac{dx}{x^2-a^2} =\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C } \]
    Exam tip: For a denominator of the form \(x^2-a^2\), factorise first and look for a difference of two simple fractions.
  • (II) INTEGRAL OF \(\displaystyle\frac{1}{a^2-x^2}\)
    ⭐HighlightStandard formula:
    \[ \boxed{ \int\frac{dx}{a^2-x^2} =\frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right|+C } \]
    where \(a\ne0\) and \(x\ne\pm a\).
    💡ConceptThis integral is similar to the previous one, but the order of the terms in the difference of squares is reversed. Factorising the denominator and separating the fractions gives a logarithmic result.
    🧩SolutionLet
    \[I=\int\frac{dx}{a^2-x^2}\]
    . Step 1: Factorise the denominator.
    \[a^2-x^2=(a-x)(a+x)\]
    Step 2: Decompose into simple fractions.
    \[\frac{1}{a^2-x^2}=\frac{1}{2a}\left(\frac{1}{a+x}+\frac{1}{a-x}\right)\]
    Consequently,
    \[I=\frac{1}{2a}\int\left(\frac{1}{a+x}+\frac{1}{a-x}\right)dx\]
    Step 3: Integrate. Recall that
    \[\int\frac{dx}{a+x}=\ln|a+x|+C\]
    and
    \[\int\frac{dx}{a-x}=-\ln|a-x|+C\]
    The negative sign in the second formula occurs because the derivative of \(a-x\) is \(-1\).

    Therefore,
    \[ \begin{aligned} I&=\frac{1}{2a} \left[\ln|a+x|-\ln|a-x|\right]+C\\ &=\frac{1}{2a} \ln\left|\frac{a+x}{a-x}\right|+C \end{aligned} \]
    Hence,
    \[\boxed{\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right|+C}\]
  • (III) INTEGRAL OF \(\displaystyle\frac{1}{x^2+a^2}\)
    ⭐HighlightStandard formula:
    \[\boxed{\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C}\]
    for \(a>0\). The formula can also be written with \(|a|\) in the denominator when \(a\ne0\).
    💡ConceptThis integral is evaluated using the derivative of the inverse tangent function. The key identity is
    \[\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^2}\]
    Scaling the variable by \(a\) allows the denominator to be converted into the form \(1+t^2\).
    🧩SolutionLet
    \[I=\int\frac{dx}{x^2+a^2}\]
    . Step 1: Put
    \[x=a\tan\theta\]
    Differentiating,
    \[dx=a\sec^2\theta\,d\theta\]
    Also,
    \[x^2+a^2=a^2\tan^2\theta+a^2\]
    \[x^2+a^2=a^2(1+\tan^2\theta)=a^2\sec^2\theta\]
    Step 2: Substitute into the integral.
    \[ \begin{aligned} I&=\int\frac{a\sec^2\theta\,d\theta} {a^2\sec^2\theta}\\ &=\frac{1}{a}\int d\theta \end{aligned} \]
    Step 3: Integrate.
    \[I=\frac{\theta}{a}+C\]
    Since \(x=a\tan\theta\), we have
    \[\theta=\tan^{-1}\left(\frac{x}{a}\right)\]
    Therefore,
    \[\boxed{\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C}\]
    Verification:
    \[ \begin{aligned} \frac{d}{dx}\left[ \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) \right] &=\frac{1}{a}\cdot \frac{1/a}{1+x^2/a^2}\\ &=\frac{1}{x^2+a^2} \end{aligned} \]
    The derivative matches the original integrand.
  • (IV) INTEGRAL OF \(\displaystyle\frac{1}{\sqrt{x^2-a^2}}\)
    ⭐HighlightStandard formula:
    \[\boxed{\int\frac{dx}{\sqrt{x^2-a^2}}=\ln\left|x+\sqrt{x^2-a^2}\right|+C}\]
    for \(a>0\) and \(x^2>a^2\). The integrand is real where \(x^2-a^2>0\).
    💡ConceptThis integral involves the square root of a difference of squares. It can be evaluated using trigonometric substitution or by recognising the derivative of a logarithmic expression. The logarithmic formula is especially useful for future exercises.
    🧩solutionLet
    \[I=\int\frac{dx}{\sqrt{x^2-a^2}}\]
    Step 1: Use trigonometric substitution.

    For the region \(x\ge a\), put
    \[x=a\sec\theta\]
    Then
    \[dx=a\sec\theta\tan\theta\,d\theta\]
    and
    \[ \begin{aligned} \sqrt{x^2-a^2} &=\sqrt{a^2\sec^2\theta-a^2}\\ &=a\sqrt{\sec^2\theta-1}\\ &=a\tan\theta \end{aligned} \]
    Here, the angle is chosen in a range where the relevant trigonometric quantities have the required signs. Step 2: Substitute.
    \[ \begin{aligned} I&=\int \frac{a\sec\theta\tan\theta\,d\theta} {a\tan\theta}\\ &=\int\sec\theta\,d\theta \end{aligned} \]
    Step 3: Apply the standard integral.
    \[ \int\sec\theta\,d\theta =\ln|\sec\theta+\tan\theta|+C \]
    Therefore,
    \[I=\ln|\sec\theta+\tan\theta|+C\]
    Step 4: Return to \(x\).

    Since \(x=a\sec\theta\),
    \[\sec\theta=\frac{x}{a}\]
    Also,
    \[\tan\theta=\frac{\sqrt{x^2-a^2}}{a}\]
    Thus,
    \[ \begin{aligned} I&=\ln\left| \frac{x}{a}+\frac{\sqrt{x^2-a^2}}{a} \right|+C\\ &=\ln\left| \frac{x+\sqrt{x^2-a^2}}{a} \right|+C \end{aligned} \]
    Since \(a\) is a non-zero constant, \(-\ln a\) can be absorbed into the integration constant for \(a>0\). Hence,
    \[ \boxed{ I=\ln\left|x+\sqrt{x^2-a^2}\right|+C } \]
    📌Important NoteThe original expression in this part is \(\frac{1}{\sqrt{x^2-a^2}}\), not \(\frac{1}{x^2-a^2}\). These are different integrals and have different standard results.
  • (V) INTEGRAL OF \(\displaystyle\frac{1}{\sqrt{a^2-x^2}}\)
    ⭐HighlightStandard formula:
    \[\boxed{\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)+C}\]
    for \(a>0\) and \(|x| < a\).
    💡ConceptThis integral is directly connected to the derivative of the inverse sine function:
    \[\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}\]
    By substituting \(x=a\sin\theta\), the square root simplifies using \(1-\sin^2\theta=\cos^2\theta\).
    🧩solutionLet
    \[I=\int\frac{dx}{\sqrt{a^2-x^2}}\]
    Step 1: Substitute.
    \[x=a\sin\theta\]
    Differentiating,
    \[dx=a\cos\theta\,d\theta\]
    Also,
    \[ \begin{aligned} \sqrt{a^2-x^2} &=\sqrt{a^2-a^2\sin^2\theta}\\ &=a\sqrt{1-\sin^2\theta}\\ &=a\cos\theta \end{aligned} \]
    Choose \(\theta\) in the principal range so that \(\cos\theta\ge0\). Step 2: Substitute into the integral.
    \[ \begin{aligned} I&=\int\frac{a\cos\theta\,d\theta}{a\cos\theta}\\ &=\int d\theta\\ &=\theta+C \end{aligned} \]
    Step 3: Return to \(x\).

    Since \(x=a\sin\theta\),
    \[\theta=\sin^{-1}\left(\frac{x}{a}\right)\]
    Therefore,
    \[\boxed{\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)+C}\]
  • (VI) INTEGRAL OF \(\displaystyle\frac{1}{\sqrt{x^2+a^2}}\)
    ⭐HighlightStandard formula:
    \[\boxed{\int\frac{dx}{\sqrt{x^2+a^2}}=\ln\left|x+\sqrt{x^2+a^2}\right|+C}\]
    for \(a\ne0\). The square root is real for all real \(x\).
    💡ConceptThe square root contains a sum of squares. A trigonometric substitution using \(x=a\tan\theta\) simplifies the expression because \(1+\tan^2\theta=\sec^2\theta\).
    🧩solutionLet
    \[I=\int\frac{dx}{\sqrt{x^2+a^2}}\]
    For simplicity, take \(a>0\) and put
    \[x=a\tan\theta\]
    Then
    \[dx=a\sec^2\theta\,d\theta\]
    and
    \[ \begin{aligned} \sqrt{x^2+a^2} &=\sqrt{a^2\tan^2\theta+a^2}\\ &=a\sqrt{1+\tan^2\theta}\\ &=a\sec\theta \end{aligned} \]
    Choose an angle range in which \(\sec\theta\) is positive.

    Step 1: Transform the integral.
    \[ \begin{aligned} I&=\int\frac{a\sec^2\theta\,d\theta}{a\sec\theta}\\ &=\int\sec\theta\,d\theta \end{aligned} \]
    Step 2: Integrate.
    \[I=\ln|\sec\theta+\tan\theta|+C\]
    Step 3: Express the result in terms of \(x\).
    \[\tan\theta=\frac{x}{a}\]
    and
    \[\sec\theta=\frac{\sqrt{x^2+a^2}}{a}\]
    Therefore,
    \[ \begin{aligned} I&=\ln\left| \frac{\sqrt{x^2+a^2}}{a}+\frac{x}{a} \right|+C\\ &=\ln\left| \frac{x+\sqrt{x^2+a^2}}{a} \right|+C \end{aligned} \]
    Since \(a\) is constant, the term \(-\ln a\) can be absorbed into \(C\). Thus,
    \[\boxed{\int\frac{dx}{\sqrt{x^2+a^2}}=\ln\left|x+\sqrt{x^2+a^2}\right|+C}\]
    Verification Differentiate the logarithmic expression:
    \[ \begin{aligned} \frac{d}{dx}\ln\left|x+\sqrt{x^2+a^2}\right| &=\frac{1+\frac{x}{\sqrt{x^2+a^2}}} {x+\sqrt{x^2+a^2}}\\ &=\frac{1}{\sqrt{x^2+a^2}} \end{aligned} \]
    The formula is verified.
  • (VII) INTEGRAL OF \(\displaystyle\frac{1}{ax^2+bx+c}\)
    ⭐HighlightGeneral form:
    \[I=\int\frac{dx}{ax^2+bx+c},\qquad a\ne0\]
    💡ConceptUnlike the preceding examples, the answer depends on the discriminant \(b^2-4ac\). Completing the square reduces the denominator to a standard quadratic form.
    🧩solutionStep 1: Complete the square
    \[ \begin{aligned} ax^2+bx+c &=a\left[x^2+\frac{b}{a}x+\frac{c}{a}\right]\\ &=a\left[ \left(x+\frac{b}{2a}\right)^2 +\frac{4ac-b^2}{4a^2} \right] \end{aligned} \]
    Put
    \[t=x+\frac{b}{2a}\]
    Then \(dt=dx\). Define the discriminant
    \[\Delta=b^2-4ac\]
    The integral becomes
    \[ I=\frac{1}{a}\int \frac{dt}{t^2-\frac{\Delta}{4a^2}} \]
    Now consider three cases.
    Case 1: \(\Delta>0\)
    Let \(k=\frac{\sqrt{\Delta}}{2|a|}\). The denominator has the form \(t^2-k^2\). Its integral is logarithmic:
    \[\int\frac{dt}{t^2-k^2}=\frac{1}{2k}\ln\left|\frac{t-k}{t+k}\right|+C\]
    Equivalently, a compact formula directly in the original coefficients is
    \[ \boxed{ \int\frac{dx}{ax^2+bx+c} =\frac{1}{\sqrt{\Delta}} \ln\left| \frac{2ax+b-\sqrt{\Delta}} {2ax+b+\sqrt{\Delta}} \right|+C } \]
    where \(\Delta=b^2-4ac>0\).
    Case 2: \(\Delta<0\)
    In this case, \(4ac-b^2>0\), so the completed-square denominator has the form \(t^2+k^2\). Use the standard inverse-tangent formula.
    \[\boxed{\int\frac{dx}{ax^2+bx+c}=\frac{2}{\sqrt{4ac-b^2}}\tan^{-1}\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)+C}\]
    This formula applies when \(a\ne0\) and \(4ac-b^2>0\).
    Case 3: \(\Delta=0\)
    The quadratic has a repeated root and can be written as
    \[ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2\]
    Therefore,
    \[ \begin{aligned} I&=\frac{1}{a}\int \frac{dx}{\left(x+\frac{b}{2a}\right)^2}\\ &=-\frac{1}{a\left(x+\frac{b}{2a}\right)}+C \end{aligned} \]
    Hence,
    \[\boxed{\int\frac{dx}{ax^2+bx+c}=-\frac{2}{2ax+b}+C}\]
    when \(\Delta=0\).
    ⭐HighlightExam strategy: For a general quadratic denominator, complete the square and inspect the discriminant. A positive discriminant gives a logarithmic form, a negative discriminant gives an inverse-tangent form, and a zero discriminant gives a reciprocal form.
  • (VIII) INTEGRAL OF \(\displaystyle\frac{1}{\sqrt{ax^2+bx+c}}\)
    🧩solutionConsider the general integral
    \[I=\int\frac{dx}{\sqrt{ax^2+bx+c}}\]
    For \(a\ne0\), complete the square:
    \[ ax^2+bx+c =a\left(x+\frac{b}{2a}\right)^2 +\frac{4ac-b^2}{4a} \]
    Let
    \[t=x+\frac{b}{2a}\]
    Then \(dt=dx\), and the integral becomes a standard form involving \(\sqrt{at^2+d}\), where \(d=\frac{4ac-b^2}{4a}\).

    For the important case \(a>0\), write
    \[ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}\]
    When the expression under the square root is positive, the standard logarithmic form is
    \[ \boxed{ \int\frac{dx}{\sqrt{ax^2+bx+c}} =\frac{1}{\sqrt a} \ln\left| 2\sqrt a\sqrt{ax^2+bx+c}+2ax+b \right|+C } \]
    for \(a>0\), on intervals where the radicand is positive. This result can be obtained by completing the square and reducing the integral to \(\int\frac{dt}{\sqrt{t^2+k^2}}\), or by differentiating the logarithmic expression to verify it.
  • (IX) INTEGRAL OF \(\displaystyle\frac{px+q}{ax^2+bx+c}\)
    💡ConceptThe numerator is linear, while the derivative of the quadratic denominator is also linear:
    \[\frac{d}{dx}(ax^2+bx+c)=2ax+b\]
    Therefore, express the numerator as a constant multiple of the derivative of the denominator plus a constant remainder.
    🧩solutionConsider
    \[I=\int\frac{px+q}{ax^2+bx+c}\,dx\]
    where \(p,q,a,b,c\) are constants and \(a\ne0\).

    Step 1: Express the numerator suitably

    Choose constants \(A\) and \(B\) such that
    \[px+q=A(2ax+b)+B\]
    Comparing coefficients of \(x\),
    \[p=2aA\]
    Therefore,
    \[A=\frac{p}{2a}\]
    Comparing the constant terms,
    \[q=Ab+B\]
    Hence,
    \[B=q-\frac{pb}{2a}\]
    Thus,
    \[px+q=\frac{p}{2a}(2ax+b)+\left(q-\frac{pb}{2a}\right)\]
    Step 2: Split the integral

    Substituting this expression into the integral,
    \[ \begin{aligned} I &=\frac{p}{2a}\int \frac{2ax+b}{ax^2+bx+c}\,dx\\ &\quad+\left(q-\frac{pb}{2a}\right) \int\frac{dx}{ax^2+bx+c} \end{aligned} \]
    The first integral is logarithmic because its numerator is the derivative of its denominator. The second is the standard quadratic-denominator integral discussed in part (vii). Therefore,
    \[ \boxed{ \begin{aligned} I&=\frac{p}{2a}\ln|ax^2+bx+c|\\ &\quad+\left(q-\frac{pb}{2a}\right) \int\frac{dx}{ax^2+bx+c} \end{aligned} } \]
    The remaining integral is evaluated according to the sign of \(b^2-4ac\), as explained in part (vii).
  • (X) INTEGRAL OF \(\displaystyle\frac{px+q}{\sqrt{ax^2+bx+c}}\)
    🧩solutionConsider
    \[I=\int\frac{px+q}{\sqrt{ax^2+bx+c}}\,dx\]
    The numerator is linear, and the derivative of the quadratic expression under the square root is also linear. This makes it possible to split the numerator into a derivative term and a constant remainder. Step 1: Express the numerator in terms of the derivative

    Since
    \[\frac{d}{dx}(ax^2+bx+c)=2ax+b\]
    write
    \[px+q=A(2ax+b)+B\]
    As in the preceding part,
    \[A=\frac{p}{2a},\quad B=q-\frac{pb}{2a}\]
    Step 2: Split the integral
    \[ \begin{aligned} I&=\frac{p}{2a} \int\frac{2ax+b}{\sqrt{ax^2+bx+c}}\,dx\\ &\quad+\left(q-\frac{pb}{2a}\right) \int\frac{dx}{\sqrt{ax^2+bx+c}} \end{aligned} \]
    For the first integral, put
    \[u=ax^2+bx+c\]
    then
    \[du=(2ax+b)\,dx\]
    Therefore,
    \[ \begin{aligned} \int\frac{2ax+b}{\sqrt{ax^2+bx+c}}\,dx &=\int\frac{du}{\sqrt u}\\ &=2\sqrt u+C\\ &=2\sqrt{ax^2+bx+c}+C \end{aligned} \]
    Consequently,
    \[ \boxed{ \begin{aligned} I&=\frac{p}{a}\sqrt{ax^2+bx+c}\\ &\quad+\left(q-\frac{pb}{2a}\right) \int\frac{dx}{\sqrt{ax^2+bx+c}} \end{aligned} } \]
    The remaining integral is the standard form given in part (viii), where its domain and coefficient conditions are satisfied.
📝 Summary of Standard Integral Forms
❌ Common Mistakes
  • Confusing \(x^2-a^2\) with \(\sqrt{x^2-a^2}\). The former produces a logarithmic ratio, while the latter produces a logarithm involving a square root.
  • Reversing the logarithmic ratio in \(\int\frac{dx}{a^2-x^2}\) without changing the sign.
  • Forgetting the chain-rule factor when differentiating or integrating a scaled variable.
  • Using the inverse-tangent formula for \(x^2-a^2\) instead of \(x^2+a^2\).
  • Failing to consider the discriminant when integrating a general quadratic denominator.
  • Ignoring the domain on which a square root is real or a denominator is non-zero.
🔑 Key Takeaway
∫

Example 7

📋
Table of Contents
4 sections
❓ Question
Evaluate the following integrals using the standard formulas for integrals involving quadratic expressions and square roots.
  1. \(\quad \int\frac{dx}{x^2-16}\)
  2. \(\quad \int\frac{dx}{\sqrt{2x-x^2}}\)
🧩 Solution
Part (i)
\(\displaystyle\int\frac{dx}{x^2-16}\)
  1. 💡
    Concept
    The denominator \(x^2-16\) is a difference of two squares. We can compare it with the standard form \(x^2-a^2\) and directly apply the corresponding logarithmic integration formula.

    The standard result is
    \[\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C\]
    where \(a\ne0\) and \(x\ne\pm a\).
    🗺️ Roadmap
    1

    Compare the denominator with \(x^2-a^2\).

    2

    Determine the value of \(a\).

    3

    Substitute the value into the standard formula.

    4

    Simplify the coefficient and write the final answer.

  2. Write the given integral.
    \[I=\int\frac{dx}{x^2-16}\]
  3. Compare with the standard form.
    \[x^2-a^2=x^2-16\]
  4. Therefore,
    \[a^2=16\]
  5. Taking \(a=4\), we have
    \[I=\int\frac{dx}{x^2-4^2}\]
  6. Apply the standard formula.
    \[ \begin{aligned} I&=\frac{1}{2(4)} \ln\left|\frac{x-4}{x+4}\right|+C\\ &=\frac{1}{8} \ln\left|\frac{x-4}{x+4}\right|+C \end{aligned} \]
  7. Hence, the final answer is
    \[\boxed{\int\frac{dx}{x^2-16}=\frac{1}{8}\ln\left|\frac{x-4}{x+4}\right|+C}\]
  8. Verification: Differentiate the result:
    \[ \begin{aligned} \frac{d}{dx}\left[ \frac18\ln\left|\frac{x-4}{x+4}\right| \right] &=\frac18\left(\frac{1}{x-4}-\frac{1}{x+4}\right)\\ &=\frac18\left(\frac{8}{x^2-16}\right)\\ &=\frac{1}{x^2-16} \end{aligned} \]
    📝 The derivative matches the original integrand, so the answer is correct.
  9. Domain
    The integrand is undefined at \(x=4\) and \(x=-4\). The antiderivative applies on intervals that do not cross these points.
Part (ii)
\(\displaystyle\int\frac{dx}{\sqrt{2x-x^2}}\)
  1. 💡
    Concept
    The expression under the square root is a quadratic polynomial. By completing the square, we can transform it into the standard form \(a^2-(x-h)^2\), which is integrated using the inverse-sine formula.

    The relevant standard result is
    \[ \int\frac{du}{\sqrt{a^2-u^2}} =\sin^{-1}\left(\frac{u}{a}\right)+C \]
    for \(a>0\) and \(|u| < a\)
    🗺️ Roadmap
    1

    Complete the square in \(2x-x^2\).

    2

    Express the denominator as \(\sqrt{1-(x-1)^2}\).

    3

    Use the substitution \(u=x-1\).

    4

    Apply the standard inverse-sine integral formula.

  2. Write the given integral.
    \[I=\int\frac{dx}{\sqrt{2x-x^2}}\]
  3. Complete the square. (Rearrange the expression inside the square root:)
    \[2x-x^2=-(x^2-2x)\]
  4. Adding and subtracting \(1\),
    \[ \begin{aligned} 2x-x^2 &=1-(x^2-2x+1)\\ &=1-(x-1)^2 \end{aligned} \]
  5. Therefore,
    \[I=\int\frac{dx}{\sqrt{1-(x-1)^2}}\]
  6. Substitute
    \[u=x-1\]
  7. Differentiating,
    \[du=dx\]
  8. The integral becomes
    \[I=\int\frac{du}{\sqrt{1-u^2}}\]
  9. Apply the standard formula.
    \[\int\frac{du}{\sqrt{1-u^2}}=\sin^{-1}u+C\]
  10. Hence,
    \[I=\sin^{-1}u+C\]
  11. Substitute \(u=x-1\)
    \[\boxed{\int\frac{dx}{\sqrt{2x-x^2}}=\sin^{-1}(x-1)+C}\]
  12. Verification: Differentiate the result:
    \[ \begin{aligned} \frac{d}{dx}\sin^{-1}(x-1) &=\frac{1}{\sqrt{1-(x-1)^2}}\\ &=\frac{1}{\sqrt{2x-x^2}} \end{aligned} \]
    📝 The derivative matches the original integrand, confirming the answer.
  13. Domain
    For the integrand to be real and finite, \(2x-x^2>0\), which gives \(0
📌
Comparison of the Two Integrals
Integral Technique Result type
\(\int\frac{dx}{x^2-16}\) Difference of squares and standard logarithmic formula Logarithmic
\(\int\frac{dx}{\sqrt{2x-x^2}}\) Completing the square and inverse-sine formula Inverse trigonometric
⚡ Exam Tip
🔑 Key Takeaway
∫

Example 8

📋
Table of Contents
4 sections
❓ Question
Find the following integrals :
  1. \(\quad \displaystyle \int \frac{dx}{x^2-6x+13}\)
  2. \(\quad \displaystyle \int \frac{dx}{3x^2+13x-10}\)
  3. \(\quad \displaystyle \int \frac{dx}{5x^2-2x}\)
🧩 Solution
Part (i)
\(\displaystyle\int \frac{dx}{x^2-6x+13}\)
  1. 💡
    Concept
    When the denominator is a quadratic expression with no real roots and can be written in the form \((x-h)^2+a^2\), we use the standard formula
    \[\int \frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C\]
    🗺️ Roadmap
    1

    Complete the square in the denominator.

    2

    Substitute \(t=x-3\).

    3

    Apply the standard inverse-tangent formula.

    4

    Substitute back and verify by differentiation.

  2. We have
    \[I=\int \frac{dx}{x^2-6x+13}\]
  3. Completing the square,
    \[ \begin{aligned} x^2-6x+13 &=x^2-6x+9+4\\ &=(x-3)^2+2^2 \end{aligned} \]
  4. Therefore,
    \[I=\int \frac{dx}{(x-3)^2+2^2}\]
  5. Let
    \[t=x-3,\quad dx=dt\]
  6. Then
    \[I=\int \frac{dt}{t^2+2^2}\]
  7. Using the standard formula,
    \[ \begin{aligned} I &=\frac{1}{2}\tan^{-1}\left(\frac{t}{2}\right)+C\\ &=\frac{1}{2}\tan^{-1}\left(\frac{x-3}{2}\right)+C \end{aligned} \]
  8. Verification Differentiating the result,
    \[ \begin{aligned} \frac{d}{dx}\left[ \frac12\tan^{-1}\left(\frac{x-3}{2}\right) \right] &=\frac12\cdot \frac{\frac12}{1+\left(\frac{x-3}{2}\right)^2}\\ &=\frac{1}{(x-3)^2+4}\\ &=\frac{1}{x^2-6x+13} \end{aligned} \]
    📝 This agrees with the original integrand.
Answer
\[ \boxed{ \int \frac{dx}{x^2-6x+13} =\frac12\tan^{-1}\left(\frac{x-3}{2}\right)+C } \]
Part (ii)
\(\displaystyle\int \frac{dx}{3x^2+13x-10}\)
  1. 💡
    Concept
    For a quadratic denominator with two distinct real roots, we can complete the square and apply the logarithmic formula

    \[\int \frac{dt}{t^2-a^2}=\frac{1}{2a}\ln\left|\frac{t-a}{t+a}\right|+C\]
    Here, the absolute value signs are necessary because the logarithm must be defined for real-valued integration on intervals where the integrand exists.
    🗺️ Roadmap
    1

    Complete the square correctly, keeping the factor \(3\) outside the quadratic expression.

    2

    Use the substitution \(t=x+\frac{13}{6}\).

    3

    Apply the logarithmic formula.

    4

    Simplify the logarithm and verify the result.

  2. We have
    \[I=\int \frac{dx}{3x^2+13x-10}\]
  3. Factor out \(3\) from the quadratic and linear terms:
    \[3x^2+13x-10=3\left(x^2+\frac{13}{3}x-\frac{10}{3}\right)\]
  4. Complete the square:
    \[ \begin{aligned}3x^2+13x-10&=3\left[x^2+\frac{13}{3}x+\left(\frac{13}{6}\right)^2-\left(\frac{13}{6}\right)^2-\frac{10}{3}\right]\\ &=3\left(x+\frac{13}{6}\right)^2-3\left(\frac{169}{36}+\frac{10}{3}\right)\\ &=3\left(x+\frac{13}{6}\right)^2-3\left(\frac{169+120}{36}\right)\\ &=3\left(x+\frac{13}{6}\right)^2-\frac{289}{12}\\ &=3\left(x+\frac{13}{6}\right)^2-\frac{289}{36}\cdot 3\\ &=3\left[\left(x+\frac{13}{6}\right)^2-\left(\frac{17}{6}\right)^2\right]\end{aligned}\]
  5. Hence,
    \[I=\frac13\int\frac{dx}{\left(x+\frac{13}{6}\right)^2-\left(\frac{17}{6}\right)^2}\]
  6. Let
    \[t=x+\frac{13}{6},\quad dx=dt\]
  7. Then
    \[I=\frac13\int\frac{dt}{t^2-\left(\frac{17}{6}\right)^2}\]
  8. Applying the standard logarithmic formula with \(a=\frac{17}{6}\),
    \[\begin{aligned}I&=\frac13\cdot\frac{1}{2\left(\frac{17}{6}\right)}\ln\left|\frac{t-\frac{17}{6}}{t+\frac{17}{6}}\right|+C\\ &=\frac13\cdot\frac{3}{17}\ln\left|\frac{t-\frac{17}{6}}{t+\frac{17}{6}}\right|+C\\ &=\frac1{17}\ln\left|\frac{t-\frac{17}{6}}{t+\frac{17}{6}}\right|+C\end{aligned}\]
  9. Substituting \(t=x+\frac{13}{6}\),
    \[\begin{aligned}I&=\frac1{17}\ln\left|\frac{x+\frac{13}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}}\right|+C\\ &=\frac1{17}\ln\left|\frac{x-\frac23}{x+5}\right|+C\\ &=\frac1{17}\ln\left|\frac{3x-2}{3(x+5)}\right|+C\end{aligned}\]
    📝 The constant term \(-\frac1{17}\ln 3\) can be absorbed into the arbitrary constant of integration.
  10. Verification
    Let
    \[F(x)=\frac1{17}\ln\left|\frac{3x-2}{x+5}\right|\]
    Using the logarithmic differentiation rule,
    \[ \begin{aligned}F'(x)&=\frac1{17}\left(\frac{3}{3x-2}-\frac{1}{x+5}\right)\\ &=\frac1{17}\left(\frac{3(x+5)-(3x-2)}{(3x-2)(x+5)}\right)\\ &=\frac1{17}\cdot\frac{17}{(3x-2)(x+5)}\\ &=\frac{1}{(3x-2)(x+5)}\\ &=\frac{1}{3x^2+13x-10} \end{aligned} \]
    📝 The derivative matches the original integrand.
Answer
\[\boxed{\int \frac{dx}{3x^2+13x-10}=\frac1{17}\ln\left|\frac{3x-2}{x+5}\right|+C}\]
Part (iii)
\(\displaystyle\int \frac{dx}{5x^2-2x}\)
  1. 💡
    Concept
    When a quadratic denominator has a common factor involving \(x\), factorisation often makes integration easier than completing the square. We use

    \[\int\frac{dx}{x(x-a)}=\frac1a\ln\left|\frac{x-a}{x}\right|+C,\quad a\ne0\]
    for intervals where the integrand is defined.
    🗺️ Roadmap
    1

    Factor \(5x^2-2x\).

    2

    Express the rational function using partial fractions.

    3

    Integrate each term using the logarithmic formula.

    4

    Verify by differentiation.

  2. Factor the denominator:
    \[5x^2-2x=x(5x-2)\]
  3. Therefore,
    \[I=\int\frac{dx}{x(5x-2)}\]
  4. Resolve into partial fractions:
    \[\frac{1}{x(5x-2)}=\frac{A}{x}+\frac{B}{5x-2}\]
  5. Multiplying both sides by \(x(5x-2)\),
    \[1=A(5x-2)+Bx\]
  6. Comparing coefficients,
    \[5A+B=0,\quad -2A=1\]
  7. Thus,
    \[A=-\frac12,\qquad B=\frac52\]
  8. Hence,
    \[\begin{aligned}I&=\int\left(-\frac{1}{2x}+\frac{5}{2(5x-2)}\right)dx\\ &=-\frac12\int\frac{dx}{x}+\frac52\int\frac{dx}{5x-2}\end{aligned}\]
  9. Using
    \[\int\frac{dx}{x}=\ln|x|+C\]
  10. and the substitution \(u=5x-2\), \(du=5\,dx\), we obtain
    \[\int\frac{dx}{5x-2}=\frac15\ln|5x-2|+C\]
  11. Therefore,
    \[\begin{aligned}I&=-\frac12\ln|x|+\frac52\cdot\frac15\ln|5x-2|+C\\ &=-\frac12\ln|x|+\frac12\ln|5x-2|+C\\ &=\frac12\ln\left|\frac{5x-2}{x}\right|+C\end{aligned}\]
  12. Verification Differentiating the result,
    \[\begin{aligned}\frac{d}{dx}\left[\frac12\ln\left|\frac{5x-2}{x}\right|\right]&=\frac12\left(\frac5{5x-2}-\frac1x\right)\\ &=\frac12\left(\frac{5x-(5x-2)}{x(5x-2)}\right)\\ &=\frac{1}{x(5x-2)}\\ &=\frac{1}{5x^2-2x}\end{aligned} \]
    📝 This confirms the answer.
Answer
\[\boxed{\int\frac{dx}{5x^2-2x}=\frac12\ln\left|\frac{5x-2}{x}\right|+C}\]
⚡ Exam Tip
🔑 Key Takeaway
∫

Example 9

📋
Table of Contents
6 sections
❓ Question
Find the integral \(\displaystyle\int \frac{x+2}{2x^2+6x+5}\,dx\)
💡 Concept
🗺️ Roadmap

  1. Express \(x+2\) in terms of the derivative of \(2x^2+6x+5\).

  2. Separate the integral into two simpler integrals.

  3. Evaluate the first integral using the logarithmic formula.

  4. Complete the square in the second integral and apply the inverse-tangent formula.

  5. Combine the results and verify by differentiation.

🧩 Solution
  1. Express the Numerator in Terms of the Derivative of the Denominator Let
    \[D(x)=2x^2+6x+5\]
  2. Then
    \[D'(x)=4x+6\]
  3. Assume that
    \[x+2=A(4x+6)+B\]
  4. Expanding,
    \[x+2=4Ax+6A+B\]
  5. Comparing coefficients of \(x\),
    \[4A=1\implies A=\frac14\]
  6. Comparing constant terms,
    \[6A+B=2\]
  7. Substituting \(A=\frac14\),
    \[\begin{aligned}\frac32+B&=2\\ B&=\frac12\end{aligned}\]
  8. Therefore,
    \[x+2=\frac14(4x+6)+\frac12\]
  9. Split the Integral Substituting the expression for the numerator,
    \[\begin{aligned}I&=\int\frac{\frac14(4x+6)+\frac12}{2x^2+6x+5}\,dx\\ &=\frac14\int\frac{4x+6}{2x^2+6x+5}\,dx+\frac12\int\frac{dx}{2x^2+6x+5}\end{aligned}\]
  10. Let
    \[I_1=\frac14\int\frac{4x+6}{2x^2+6x+5}\,dx\]
    and
    \[I_2=\frac12\int\frac{dx}{2x^2+6x+5}\]
  11. Thus,
    \[I=I_1+I_2\]
  12. Evaluate \(I_1\) Let
    \[t=2x^2+6x+5\]
  13. Differentiating,
    \[dt=(4x+6)\,dx\]
  14. Therefore,
    \[\begin{aligned}I_1&=\frac14\int\frac{dt}{t}\\ &=\frac14\ln|t|+C\\ &=\frac14\ln|2x^2+6x+5|+C\end{aligned}\]
  15. Notice that
    \[2x^2+6x+5=2\left(x+\frac32\right)^2+\frac12>0\]
  16. for every real \(x\). Hence, the absolute value signs can be omitted:
    \[\boxed{I_1=\frac14\ln(2x^2+6x+5)}\]
  17. Evaluate \(I_2\) by Completing the Square We have
    \[I_2=\frac12\int\frac{dx}{2x^2+6x+5}\]
  18. Factor out \(2\) from the denominator:
    \[I_2=\frac14\int\frac{dx}{x^2+3x+\frac52}\]
  19. Complete the square:
    \[\begin{aligned}x^2+3x+\frac52&=x^2+3x+\frac94-\frac94+\frac52\\ &=\left(x+\frac32\right)^2-\frac94+\frac{10}{4}\\ &=\left(x+\frac32\right)^2+\frac14\end{aligned}\]
  20. Therefore,
    \[I_2=\frac14\int\frac{dx}{\left(x+\frac32\right)^2+\left(\frac12\right)^2}\]
  21. Use the standard formula
    \[\int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac xa\right)+C\]
  22. Let
    \[t=x+\frac32,\qquad dt=dx\]
  23. Then
    \[\begin{aligned}I_2&=\frac14\int\frac{dt}{t^2+\left(\frac12\right)^2}\\ &=\frac14\cdot 2\tan^{-1}\left(\frac{t}{\frac12}\right)+C\\ &=\frac12\tan^{-1}(2t)+C\\ &=\frac12\tan^{-1}\left(2x+3\right)+C\end{aligned}\]
  24. Hence,
    \[\boxed{I_2=\frac12\tan^{-1}(2x+3)}\]
  25. Combine the Results Since \(I=I_1+I_2\), we obtain
    \[\boxed{\begin{aligned}\int\frac{x+2}{2x^2+6x+5}\,dx&=\frac14\ln(2x^2+6x+5)\\ &\quad+\frac12\tan^{-1}(2x+3)+C\end{aligned}}\]
  26. Verification by Differentiation Let
    \[F(x)=\frac14\ln(2x^2+6x+5)+\frac12\tan^{-1}(2x+3)\]
  27. Differentiating the logarithmic term,
    \[\frac{d}{dx}\left[\frac14\ln(2x^2+6x+5)\right]=\frac{4x+6}{4(2x^2+6x+5)}\]
  28. Differentiating the inverse-tangent term,
    \[\begin{aligned}\frac{d}{dx}\left[\frac12\tan^{-1}(2x+3)\right]&=\frac12\cdot\frac{2}{1+(2x+3)^2}\\ &=\frac{1}{1+(2x+3)^2}\\ &=\frac{1}{4x^2+12x+10}\\ &=\frac{1}{2(2x^2+6x+5)}\end{aligned}\]
  29. Adding the derivatives,
    \[\begin{aligned}F'(x)&=\frac{4x+6}{4(2x^2+6x+5)}+\frac{1}{2(2x^2+6x+5)}\\ &=\frac{4x+6+2}{4(2x^2+6x+5)}\\ &=\frac{4x+8}{4(2x^2+6x+5)}\\ &=\frac{x+2}{2x^2+6x+5}\end{aligned}\]
    📝 This is exactly the original integrand. Therefore, the answer is verified.
❌ Common Mistakes
Exam Tips and Common Mistakes
  • Identify the derivative: the derivative of \(2x^2+6x+5\) is \(4x+6\), not \(4x+3\).
  • Split the numerator correctly: \(x+2=\frac14(4x+6)+\frac12\).
  • Keep the factors outside the integral: the second integral carries a factor of \(\frac12\), which combines with the factor \(\frac12\) obtained by factoring \(2\) from the denominator.
  • Complete the square accurately: \(x^2+3x+\frac52=\left(x+\frac32\right)^2+\frac14\).
  • Use the correct inverse-tangent argument: \(\frac{x+\frac32}{\frac12}=2x+3\).
  • Verify by differentiation: this is especially useful for catching missing constant factors.
🔑 Key Takeaway
∫

Integration by Partial Fractions

📋
Table of Contents
3 sections
📘 Definition
🗂️ Standard Forms of Partial Fraction Decomposition
  • CASE I: TWO DISTINCT LINEAR FACTORS
    If the denominator contains two distinct linear factors, then
    \[\frac{px+q}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b},\quad a\neq b\]
    Here, \(A\) and \(B\) are constants determined by comparing coefficients or substituting suitable values of \(x\).
  • CASE II: A REPEATED LINEAR FACTOR
    If the denominator contains a repeated linear factor, then
    \[\frac{px+q}{(x-a)^2}=\frac{A}{x-a}+\frac{B}{(x-a)^2}\]
    For a factor repeated \(n\) times, include every power from \(1\) to \(n\):
    \[\frac{P(x)}{(x-a)^n}=\frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+\cdots+\frac{A_n}{(x-a)^n}\]
  • CASE III: THREE DISTINCT LINEAR FACTORS
    If the denominator contains three distinct linear factors, then
    \[\frac{px^2+qx+r}{(x-a)(x-b)(x-c)}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}\]
    Here, the numerator of the third partial fraction must be a constant \(C\), not \(c\), because \(c\) already denotes one of the factor parameters.
  • CASE IV: ONE REPEATED LINEAR FACTOR AND ONE DISTINCT LINEAR FACTOR
    If the denominator contains a repeated factor \((x-a)^2\) and a distinct factor \((x-b)\), then
    \[\frac{px^2+qx+r}{(x-a)^2(x-b)}=\frac{A}{x-a}+\frac{B}{(x-a)^2}+\frac{C}{x-b}\]
    For higher powers, include a partial fraction for every power of the repeated factor.
  • CASE V: ONE LINEAR FACTOR AND ONE IRREDUCIBLE QUADRATIC FACTOR
    If the denominator contains a linear factor and an irreducible quadratic factor, then
    \[\frac{px^2+qx+r}{(x-a)(x^2+bx+c)}=\frac{A}{x-a}+\frac{Bx+C}{x^2+bx+c}\]
    The quadratic \(x^2+bx+c\) is irreducible over the real numbers when
    \[b^2-4c<0\]
    The numerator over an irreducible quadratic factor must be linear, \(Bx+C\), rather than just a constant. This allows the decomposition to represent a general proper rational function.
📜 Important Rules
📜 Rules
  • Every distinct linear factor contributes a constant numerator.
  • A repeated linear factor requires a term for each power of that factor.
  • An irreducible quadratic factor requires a linear numerator.
  • If the numerator degree is greater than or equal to the denominator degree, perform polynomial division first.
  • After finding the constants, integrate each partial fraction separately.
∫

Example 10

📋
Table of Contents
6 sections
❓ Question
Evaluate the integral
\[\displaystyle\int \frac{dx}{(x+1)(x+2)}\]
💡 Concept
🗺️ Roadmap

  1. Express the rational function as a sum of two partial fractions.

  2. Determine the unknown constants by comparing coefficients.

  3. Integrate the resulting terms separately.

  4. Combine the logarithms and verify the answer by differentiation.

🧩 Solution
  1. Resolve into Partial Fractions Assume that
    \[\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}\]
  2. Multiplying both sides by \((x+1)(x+2)\), we obtain
    \[1=A(x+2)+B(x+1)\]
  3. Expanding the right-hand side,
    \[1=Ax+2A+Bx+B\]
  4. Rearranging,
    \[1=(A+B)x+(2A+B)\]
  5. Comparing the coefficients of \(x\),
    \[A+B=0\]
  6. Therefore,
    \[B=-A\]
  7. Comparing the constant terms,
    \[2A+B=1\]
  8. Substituting \(B=-A\),
    \[\begin{aligned}2A-A&=1\\ A&=1\end{aligned}\]
  9. Hence,
    \[B=-1\]
  10. Thus, the partial fraction decomposition is
    \[\boxed{\frac{1}{(x+1)(x+2)}=\frac{1}{x+1}-\frac{1}{x+2}}\]
  11. Integrate Each Partial Fraction Substituting the decomposition into the original integral,
    \[\begin{aligned}I&=\int\left(\frac{1}{x+1}-\frac{1}{x+2}\right)dx\\ &=\int\frac{dx}{x+1}-\int\frac{dx}{x+2}\end{aligned}\]
  12. Using the standard logarithmic integral formula,
    \[\int\frac{dx}{x+1}=\ln|x+1|+C\]
    and
    \[\int\frac{dx}{x+2}=\ln|x+2|+C\]
  13. we obtain
    \[I=\ln|x+1|-\ln|x+2|+C\]
  14. Combine the Logarithms Using the logarithmic identity
    \[\ln|u|-\ln|v|=\ln\left|\frac{u}{v}\right|\]
  15. we get
    \[I=\ln\left|\frac{x+1}{x+2}\right|+C\]
  16. Verification by Differentiation Let
    \[F(x)=\ln|x+1|-\ln|x+2|\]
  17. Differentiating with respect to \(x\),
    \[\begin{aligned}F'(x)&=\frac{1}{x+1}-\frac{1}{x+2}\\ &=\frac{(x+2)-(x+1)}{(x+1)(x+2)}\\ &=\frac{1}{(x+1)(x+2)}\end{aligned}\]
    📝 This is the original integrand. Therefore, the result is verified.
Answer
\[\boxed{\int\frac{dx}{(x+1)(x+2)}=\ln\left|\frac{x+1}{x+2}\right|+C}\]
❌ Common Mistakes
  • Multiply both sides by the complete denominator before comparing coefficients.
  • Remember that \(A+B=0\) and \(2A+B=1\), giving \(A=1\) and \(B=-1\).
  • Retain absolute value signs in logarithmic expressions for real-valued integration.
  • Do not reverse the ratio when combining logarithms; the correct ratio is \(\frac{x+1}{x+2}\).
  • The integrand is undefined at \(x=-1\) and \(x=-2\). The antiderivative applies on intervals that do not cross these points.
🔑 Key Takeaway
∫

Example 11

📋
Table of Contents
6 sections
❓ Question
Evaluate \(\displaystyle\int\frac{x^2+1}{x^2-5x+6}\,dx\)
💡 Concept
🗺️ Roadmap
The denominator \(x^2-5x+6\) factorises into two distinct linear factors. After polynomial division, the remaining numerator is linear, so we can use partial fractions of the form

\[ \frac{5x-5}{(x-2)(x-3)} =\frac{A}{x-2}+\frac{B}{x-3} \]
🧩 Solution
  1. Factorise the Denominator Consider
    \[x^2-5x+6\]
  2. Split the middle term as \(-3x-2x\):
    \[\begin{aligned}x^2-5x+6&=x^2-3x-2x+6\\[4pt] &=x(x-3)-2(x-3)\\[4pt] &=(x-3)(x-2)\end{aligned}\]
  3. Therefore,
    \[I=\int\frac{x^2+1}{(x-2)(x-3)}\,dx\]
  4. Perform Polynomial Division Since the numerator and denominator have the same degree, write the numerator as the denominator plus a remainder:
    \[\begin{aligned}x^2+1&=(x^2-5x+6)+(5x-5)\end{aligned}\]
  5. Dividing by \(x^2-5x+6\),
    \[\frac{x^2+1}{x^2-5x+6}=1+\frac{5x-5}{x^2-5x+6}\]
  6. Hence,
    \[I=\int 1\,dx+\int\frac{5x-5}{(x-2)(x-3)}\,dx\]
  7. Resolve into Partial Fractions Assume that
    \[\frac{5x-5}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}\]
  8. Multiplying both sides by \((x-2)(x-3)\),
    \[5x-5=A(x-3)+B(x-2)\]
  9. Expanding,
    \[5x-5=Ax-3A+Bx-2B\]
  10. Collecting like terms,
    \[5x-5=(A+B)x-(3A+2B)\]
  11. Comparing coefficients of \(x\),
    \[A+B=5\]
  12. Comparing constant terms,
    \[-3A-2B=-5\]
  13. Equivalently,
    \[3A+2B=5\]
  14. From \(A+B=5\),
    \[A=5-B\]
  15. Substituting into \(3A+2B=5\),
    \[\begin{aligned}3(5-B)+2B&=5\\[4pt] 15-3B+2B&=5\\[4pt] -B&=-10\\[4pt] B&=10\end{aligned}\]
  16. Therefore,
    \[A=5-10=-5\]
  17. Integrate Each Term Thus, the partial fraction decomposition is
    \[\boxed{\frac{5x-5}{(x-2)(x-3)}=-\frac5{x-2}+\frac{10}{x-3}}\]
  18. Substituting the decomposition,
    \[I=\int\left(1-\frac5{x-2}+\frac{10}{x-3}\right)dx\]
  19. Integrating term by term,
    \[\begin{aligned}I&=\int 1\,dx-5\int\frac{dx}{x-2}+10\int\frac{dx}{x-3}\\[4pt] &=x-5\ln|x-2|+10\ln|x-3|+C\end{aligned}\]
  20. Combine the Logarithms Using the logarithmic identities,
    \[\begin{aligned}I&=x+\ln|x-3|^{10}-\ln|x-2|^5+C\\[4pt] &=x+\ln\left|\frac{(x-3)^{10}}{(x-2)^5}\right|+C\end{aligned}\]
  21. Verification by Differentiation Let
    \[F(x)=x-5\ln|x-2|+10\ln|x-3|\]
  22. Differentiating
    \[\begin{aligned}F'(x)&=1-\frac5{x-2}+\frac{10}{x-3}\\[4pt] &=1+\frac{-5(x-3)+10(x-2)}{(x-2)(x-3)}\\[4pt] &=1+\frac{-5x+15+10x-20}{(x-2)(x-3)}\\[4pt] &=1+\frac{5x-5}{(x-2)(x-3)}\\[4pt] &=\frac{(x-2)(x-3)+5x-5}{(x-2)(x-3)}\\[4pt] &=\frac{x^2-5x+6+5x-5}{x^2-5x+6}\\[4pt] &=\frac{x^2+1}{x^2-5x+6}\end{aligned}\]
    📝 This matches the original integrand, confirming the result.
  23. Domain Consideration
    The integrand is undefined at \(x=2\) and \(x=3\). Therefore, the antiderivative is valid on any interval that does not cross these points. The integration constant may differ on the separate intervals.
Answer
\[\boxed{\int\frac{x^2+1}{x^2-5x+6}\,dx=x-5\ln|x-2|+10\ln|x-3|+C}\]
❌ Common Mistakes
  • Perform polynomial division first: because the numerator and denominator have equal degree, the rational function must first be written as \(1+\frac{5x-5}{x^2-5x+6}\).
  • Factorise accurately: \(x^2-5x+6=(x-2)(x-3)\).
  • Check the signs of the constants: the correct values are \(A=-5\) and \(B=10\).
  • Use absolute values: write \(\ln|x-2|\) and \(\ln|x-3|\) for a real-valued antiderivative.
  • Verify by differentiation: this confirms the decomposition and the final result.
🔑 Key Takeaway
∫

Example 12

📋
Table of Contents
6 sections
❓ Question
Evaluate \(\displaystyle \int\frac{3x-2}{(x+1)^2(x+3)}\,dx\)
💡 Concept
🗺️ Roadmap

  1. Set up the partial fraction decomposition.

  2. Multiply by the denominator and determine \(A\), \(B\), and \(C\).

  3. Integrate the logarithmic terms.

  4. Integrate the repeated-factor term using the power rule.

  5. Verify the result by differentiation.

🧩 Solution
  1. Resolve into Partial Fractions Assume that
    \[\frac{3x-2}{(x+1)^2(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+3}\]
  2. Multiplying both sides by \((x+1)^2(x+3)\), we obtain
    \[3x-2=A(x+1)(x+3)+B(x+3)+C(x+1)^2\]
  3. Expanding each term,
    \[\begin{aligned}3x-2&=A(x^2+4x+3)+B(x+3)+C(x^2+2x+1)\\ &=Ax^2+4Ax+3A+Bx+3B+Cx^2+2Cx+C\\ &=(A+C)x^2+(4A+B+2C)x+(3A+3B+C)\end{aligned}\]
  4. Comparing coefficients of \(x^2\),
    \[A+C=0\]
  5. Therefore,
    \[A=-C\]
  6. Comparing coefficients of \(x\),
    \[4A+B+2C=3\]
  7. Substituting \(A=-C\),
    \[\begin{aligned}-4C+B+2C&=3\\ B-2C&=3\\ B&=3+2C\end{aligned}\]
  8. Comparing constant terms,
    \[3A+3B+C=-2\]
  9. Substituting \(A=-C\) and \(B=3+2C\),
    \[\begin{aligned}3(-C)+3(3+2C)+C&=-2\\ -3C+9+6C+C&=-2\\ 4C+9&=-2\\ 4C&=-11\\ C&=-\frac{11}{4}\end{aligned}\]
  10. Since \(A=-C\),
    \[A=\frac{11}{4}\]
  11. Also,
    \[\begin{aligned}B&=3+2C\\ &=3+2\left(-\frac{11}{4}\right)\\ &=3-\frac{11}{2}\\ &=-\frac52\end{aligned}\]
  12. Therefore, the partial fraction decomposition is
    \[\boxed{\frac{3x-2}{(x+1)^2(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^2}-\frac{11}{4(x+3)}}\]
  13. Integrate Each Term Substituting the decomposition into the original integral,
    \[\begin{aligned}I&=\int\left[\frac{11}{4(x+1)}-\frac{5}{2(x+1)^2}-\frac{11}{4(x+3)}\right]dx\\[5pt] &=\frac{11}{4}\int\frac{dx}{x+1}-\frac52\int\frac{dx}{(x+1)^2}-\frac{11}{4}\int\frac{dx}{x+3}\end{aligned}\]
  14. Integrate the Logarithmic Terms Using the standard formula
    \[\int\frac{dx}{x+a}=\ln|x+a|+C\]
  15. we get
    \[\frac{11}{4}\int\frac{dx}{x+1}=\frac{11}{4}\ln|x+1|\]
  16. Similarly,
    \[-\frac{11}{4}\int\frac{dx}{x+3}=-\frac{11}{4}\ln|x+3|\]
  17. Integrate the Repeated-Factor Term Consider
    \[-\frac52\int\frac{dx}{(x+1)^2}\]
  18. Let
    \[t=x+1,\quad dt=dx\]
  19. Then
    \[-\frac52\int\frac{dt}{t^2}=-\frac52\int t^{-2}\,dt\]
  20. ⭐ Using the power rule,
    \[\int t^n\,dt=\frac{t^{n+1}}{n+1}+C,\qquad n\neq-1\]
  21. we obtain
    \[\begin{aligned}-\frac52\int t^{-2}\,dt&=-\frac52\cdot\frac{t^{-1}}{-1}+C\\[5pt] &=\frac52t^{-1}+C\\[5pt] &=\frac{5}{2t}+C\\[5pt] &=\frac{5}{2(x+1)}+C\end{aligned}\]
  22. Therefore,
    \[-\frac52\int\frac{dx}{(x+1)^2}=\frac{5}{2(x+1)}+C\]
  23. Combine the Results Combining all three integrals,
    \[I=\frac{11}{4}\ln|x+1|-\frac{11}{4}\ln|x+3|+\frac{5}{2(x+1)}+C\]
  24. Using the logarithmic identity,
    \[\ln|u|-\ln|v|=\ln\left|\frac{u}{v}\right|\]
  25. we obtain the final result.
    \[\boxed{\begin{aligned}\int\frac{3x-2}{(x+1)^2(x+3)}\,dx&=\frac{11}{4}\ln\left|\frac{x+1}{x+3}\right|\\[5pt] &\quad+\frac{5}{2(x+1)}+C\end{aligned}}\]
  26. Verification by Differentiation Let
    \[F(x)=\frac{11}{4}\ln\left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}\]
  27. Differentiating,
    \[\begin{aligned}F'(x)&=\frac{11}{4}\left(\frac{1}{x+1}-\frac{1}{x+3}\right)-\frac{5}{2(x+1)^2}\\[5pt] &=\frac{11}{4}\cdot\frac{2}{(x+1)(x+3)}-\frac{5}{2(x+1)^2}\\[5pt] &=\frac{11}{2(x+1)(x+3)}-\frac{5}{2(x+1)^2}\end{aligned}\]
  28. Taking the common denominator \(2(x+1)^2(x+3)\),
    \[\begin{aligned}F'(x)&=\frac{11(x+1)-5(x+3)}{2(x+1)^2(x+3)}\\[5pt] &=\frac{11x+11-5x-15}{2(x+1)^2(x+3)}\\[5pt] &=\frac{6x-4}{2(x+1)^2(x+3)}\\[5pt] &=\frac{3x-2}{(x+1)^2(x+3)}\end{aligned}\]
    📝 This matches the original integrand, confirming the result.
  29. Domain Consideration
    The integrand is undefined at \(x=-1\) and \(x=-3\). The antiderivative is valid on any interval that does not cross these points. The integration constant may differ on separate intervals.
❌ Common Mistakes
  • Include every repeated-factor term: for \((x+1)^2\), include both \(\frac{A}{x+1}\) and \(\frac{B}{(x+1)^2}\).
  • Use consistent notation: use \(C\) for the coefficient of \(\frac{1}{x+3}\), and reserve the final \(C\) for the integration constant.
  • Compare all coefficients: equate the coefficients of \(x^2\), \(x\), and the constant terms.
  • Apply the power rule correctly: \(\int (x+1)^{-2}dx=-\frac{1}{x+1}+C\).
  • Retain absolute values: logarithmic terms should contain absolute values for real-valued integration.
🔑 Key Takeaway
∫

Example 13

📋
Table of Contents
6 sections
❓ Question
Evaluate \(displaystyle \int\frac{x^2}{(x^2+1)(x^2+4)}\,dx\)
💡 Concept
🗺️ Roadmap

  1. Write \(t=x^2\) to simplify the algebraic decomposition.

  2. Find the constants \(A\) and \(B\).

  3. Rewrite the original integrand in partial fractions.

  4. Integrate each term using the inverse-tangent formula.

  5. Verify the result by differentiation.

🧩 Solution
  1. Resolve into Partial Fractions Let \(t=x^2\). Then the rational expression becomes
    \[\frac{t}{(t+1)(t+4)}\]
  2. Assume that
    \[\frac{t}{(t+1)(t+4)}=\frac{A}{t+1}+\frac{B}{t+4}\]
  3. Multiplying both sides by \((t+1)(t+4)\),
    \[t=A(t+4)+B(t+1)\]
  4. Expanding,
    \[\begin{aligned}t&=At+4A+Bt+B\\ &=(A+B)t+(4A+B)\end{aligned}\]
  5. Comparing the coefficients of \(t\),
    \[A+B=1\]
  6. Comparing the constant terms,
    \[4A+B=0\]
  7. Subtracting the first equation from the second,
    \[\begin{aligned}(4A+B)-(A+B)&=0-1\\ 3A&=-1\\ A&=-\frac13\end{aligned}\]
  8. Substituting \(A=-\frac13\) into \(A+B=1\),
    \[\begin{aligned}-\frac13+B&=1\\ B&=1+\frac13\\ B&=\frac43\end{aligned}\]
  9. Therefore,
    \[\boxed{A=-\frac13,\qquad B=\frac43}\]
  10. Write the Partial Fraction Decomposition Replacing \(t\) with \(x^2\),
    \[\boxed{\frac{x^2}{(x^2+1)(x^2+4)}=-\frac{1}{3(x^2+1)}+\frac{4}{3(x^2+4)}}\]
  11. Integrate Each Term Substituting the decomposition into the integral,
    \[\begin{aligned}I&=-\frac13\int\frac{dx}{x^2+1}+\frac43\int\frac{dx}{x^2+4}\end{aligned}\]
  12. For the first integral, using \(a=1\),
    \[\int\frac{dx}{x^2+1}=\tan^{-1}x+C\]
  13. For the second integral, \(4=2^2\), so \(a=2\):
    \[\int\frac{dx}{x^2+4}=\frac12\tan^{-1}\left(\frac{x}{2}\right)+C\]
  14. Therefore,
    \[\begin{aligned}I&=-\frac13\tan^{-1}x+\frac43\cdot\frac12\tan^{-1}\left(\frac{x}{2}\right)+C\\ &=-\frac13\tan^{-1}x+\frac23\tan^{-1}\left(\frac{x}{2}\right)+C\end{aligned}\]
  15. Verification by Differentiation Let
    \[F(x)=-\frac13\tan^{-1}x+\frac23\tan^{-1}\left(\frac{x}{2}\right)\]
  16. Differentiating,
    \[\begin{aligned}F'(x)&=-\frac{1}{3(1+x^2)}+\frac23\cdot\frac{\frac12}{1+\left(\frac{x}{2}\right)^2}\\[5pt] &=-\frac{1}{3(x^2+1)}+\frac13\cdot\frac{1}{1+\frac{x^2}{4}}\\[5pt] &=-\frac{1}{3(x^2+1)}+\frac{4}{3(x^2+4)}\end{aligned}\]
  17. Combining the fractions,
    \[\begin{aligned}F'(x)&=\frac{-(x^2+4)+4(x^2+1)}{3(x^2+1)(x^2+4)}\\[5pt] &=\frac{-x^2-4+4x^2+4}{3(x^2+1)(x^2+4)}\\[5pt] &=\frac{3x^2}{3(x^2+1)(x^2+4)}\\[5pt] &=\frac{x^2}{(x^2+1)(x^2+4)}\end{aligned}\]
    📝 This is exactly the original integrand, confirming the answer.
Answer
\[\boxed{\int\frac{x^2}{(x^2+1)(x^2+4)}\,dx=-\frac13\tan^{-1}x+\frac23\tan^{-1}\left(\frac{x}{2}\right)+C}\]
❌ Common Mistakes
  • Keep the substitution in perspective: writing \(t=x^2\) simplifies the algebraic decomposition, but the integration variable remains \(x\).
  • Compare both coefficients: the equations are \(A+B=1\) and \(4A+B=0\), giving \(A=-\frac13\) and \(B=\frac43\).
  • Use the correct inverse-tangent formula: \(\int\frac{dx}{x^2+4}=\frac12\tan^{-1}\left(\frac{x}{2}\right)+C\).
  • Check the numerical factors: the coefficient \(\frac43\) multiplied by \(\frac12\) gives \(\frac23\).
  • Verify by differentiation: this confirms the signs and coefficients of the final answer.
🔑 Key Takeaway
∫

Exercise 14

📋
Table of Contents
6 sections
❓ Question
Evaluate \(\displaystyle\int\frac{x^2+x+1}{(x+2)(x^2+1)}\,dx\)
💡 Concept
🗺️ Roadmap

  1. Express the rational function as a sum of two partial fractions.

  2. Determine the constants \(A\), \(B\), and \(C\) by comparing coefficients.

  3. Separate the numerator of the quadratic fraction into a multiple of the derivative of \(x^2+1\) and a constant.

  4. Integrate using logarithmic and inverse-tangent formulas.

  5. Verify the final answer by differentiation.

🧩 Solution
  1. Resolve into Partial Fractions Assume that
    \[\frac{x^2+x+1}{(x+2)(x^2+1)}=\frac{A}{x+2}+\frac{Bx+C}{x^2+1}\]
  2. Multiplying both sides by \((x+2)(x^2+1)\),
    \[x^2+x+1=A(x^2+1)+(Bx+C)(x+2)\]
  3. Expanding,
    \[\begin{aligned}x^2+x+1&=Ax^2+A+Bx^2+2Bx+Cx+2C\\ &=(A+B)x^2+(2B+C)x+(A+2C)\end{aligned}\]
  4. Comparing the coefficients of \(x^2\),
    \[A+B=1\]
  5. Therefore,
    \[B=1-A\]
  6. Comparing the coefficients of \(x\),
    \[2B+C=1\]
  7. Comparing the constant terms,
    \[A+2C=1\]
  8. Determine the Constants From \(B=1-A\), the equation \(2B+C=1\) gives
    \[\begin{aligned}2(1-A)+C&=1\\ 2-2A+C&=1\\ 2A&=1+C\\ A&=\frac{1+C}{2}\end{aligned}\]
  9. Substitute this into \(A+2C=1\):
    \[\begin{aligned}\frac{1+C}{2}+2C&=1\\ 1+C+4C&=2\\ 5C&=1\\ C&=\frac15\end{aligned}\]
  10. Now,
    \[\begin{aligned}A&=\frac{1+\frac15}{2}\\ &=\frac{\frac65}{2}\\ &=\frac35\end{aligned}\]
  11. Finally,
    \[\begin{aligned}B&=1-A\\ &=1-\frac35\\ &=\frac25\end{aligned}\]
  12. Thus,
    \[\boxed{A=\frac35,\qquad B=\frac25,\qquad C=\frac15}\]
  13. Write the Partial Fraction Decomposition Substituting these values,
    \[\begin{aligned}\frac{x^2+x+1}{(x+2)(x^2+1)}&=\frac{3}{5(x+2)}+\frac{\frac25x+\frac15}{x^2+1}\\[5pt] &=\frac{3}{5(x+2)}+\frac{2x+1}{5(x^2+1)}\end{aligned}\]
  14. Therefore,
    \[I=\frac35\int\frac{dx}{x+2}+\frac15\int\frac{2x+1}{x^2+1}\,dx\]
  15. Separate the second integral:
    \[\begin{aligned}I&=\frac35\int\frac{dx}{x+2}+\frac15\int\frac{2x}{x^2+1}\,dx\\[5pt] &\quad+\frac15\int\frac{dx}{x^2+1}\end{aligned}\]
  16. Evaluate the Individual Integrals First integral
    \[\frac35\int\frac{dx}{x+2}=\frac35\ln|x+2|\]
  17. Second integral:
    \[\frac15\int\frac{2x}{x^2+1}\,dx\]
  18. Let
    \[t=x^2+1,\quad dt=2x\,dx\]
  19. Then
    \[\begin{aligned}\frac15\int\frac{2x}{x^2+1}\,dx&=\frac15\int\frac{dt}{t}\\[5pt] &=\frac15\ln|t|+C\\[5pt] &=\frac15\ln(x^2+1)+C\end{aligned}\]
    📝 Since \(x^2+1>0\) for every real \(x\), the absolute value signs are unnecessary.
  20. Third integral: Using the standard formula
    \[\int\frac{dx}{x^2+1}=\tan^{-1}x+C\]
  21. we obtain
    \[\frac15\int\frac{dx}{x^2+1}=\frac15\tan^{-1}x\]
  22. Combine the Results Adding all three integrals,
    \[\begin{aligned}I&=\frac35\ln|x+2|+\frac15\ln(x^2+1)+\frac15\tan^{-1}x+C\end{aligned}\]
  23. Verification by Differentiation Let
    \[F(x)=\frac35\ln|x+2|+\frac15\ln(x^2+1)+\frac15\tan^{-1}x\]
  24. Differentiating,
    \[\begin{aligned}F'(x)&=\frac{3}{5(x+2)}+\frac{2x}{5(x^2+1)}+\frac{1}{5(x^2+1)}\\[5pt] &=\frac{3}{5(x+2)}+\frac{2x+1}{5(x^2+1)}\end{aligned}\]
  25. Taking the common denominator \(5(x+2)(x^2+1)\),
    \[\begin{aligned}F'(x)&=\frac{3(x^2+1)+(2x+1)(x+2)}{5(x+2)(x^2+1)}\\[5pt] &=\frac{3x^2+3+2x^2+5x+2}{5(x+2)(x^2+1)}\\[5pt] &=\frac{5x^2+5x+5}{5(x+2)(x^2+1)}\\[5pt] &=\frac{x^2+x+1}{(x+2)(x^2+1)}\end{aligned}\]
    📝 This is exactly the original integrand, confirming the result.
  26. Domain Consideration
    The denominator is zero only at \(x=-2\), since \(x^2+1\) is always positive for real \(x\). The antiderivative is therefore valid on either interval \(x<-2\) or \(x>-2\), with the integration constant chosen independently on each interval.
\[\boxed{\begin{aligned}\int\frac{x^2+x+1}{(x+2)(x^2+1)}\,dx&=\frac35\ln|x+2|\\[5pt] &\quad+\frac15\ln(x^2+1)\\[5pt]&\quad+\frac15\tan^{-1}x+C\end{aligned}}\]
❌ Exam Tips and Common Mistakes
  • Use a linear numerator for the quadratic factor: write \(\frac{Bx+C}{x^2+1}\), not just \(\frac{B}{x^2+1}\).
  • Compare all three coefficients: use the coefficients of \(x^2\), \(x\), and the constant term to obtain three equations.
  • Recognise the derivative: the term \(2x\) is the derivative of \(x^2+1\), so it produces a logarithm.
  • Handle the remaining constant numerator: \(\int\frac{dx}{x^2+1}=\tan^{-1}x+C\).
  • Keep absolute values where necessary: use \(\ln|x+2|\), but \(\ln(x^2+1)\) needs no absolute value.
🔑 Key Takeaway
∫

Integration by Parts

📋
Table of Contents
9 sections
📘 Definition
📐 Derivation
🔢 Formula in Terms of Functions of \(x\)
🗒️ Standard Notation
The formula is generally written in the compact form
\[\boxed{\int u\,v\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx}\]
Equivalently, when \(dv=v(x)\,dx\) and \(du=u'(x)\,dx\), the preferred differential notation is
\[\boxed{\int u\,dv=uv-\int v\,du}\]
The differential form is often easier to apply because it clearly identifies which function is differentiated and which is integrated.
🧰 How to Choose \(u\) and \(dv\)
The success of integration by parts depends on selecting \(u\) and \(dv\) appropriately. Usually, choose \(u\) as the function that becomes simpler when differentiated, and choose \(dv\) as the part that can be integrated easily.

A commonly used guideline is the ILATE rule, which suggests the following priority for choosing \(u\):
  1.   I :    Inverse trigonometric functions, such as \(\tan^{-1}x\).
  2.   L :    Logarithmic functions, such as \(\log x\).
  3.   A :    Algebraic functions, such as \(x\) or \(x^2\).
  4.   T :    Trigonometric functions, such as \(\sin x\) or \(\cos x\).
  5.   E :    Exponential functions, such as \(e^x\).
ILATE is a useful guideline rather than an absolute rule. The best choice is the one that simplifies the remaining integral.
🛠️ Application
Integration by parts is particularly useful for integrals involving the following products:
  • Algebraic and exponential functions: \(\int xe^x\,dx\).
  • Algebraic and trigonometric functions: \(\int x\sin x\,dx\).
  • Algebraic and logarithmic functions: \(\int x\log x\,dx\).
  • Algebraic and inverse trigonometric functions: \(\int x\tan^{-1}x\,dx\).
✏️ Example
  1. Choose \(u\) and \(dv\). Using the ILATE guideline, choose
    \[u=x,\quad dv=e^x\,dx\]
  2. Then
    \[du=dx,\quad v=e^x\]
  3. Apply the formula
    \[\begin{aligned}I&=uv-\int v\,du\\ &=xe^x-\int e^x\,dx\\ &=xe^x-e^x+C\end{aligned}\]
  4. Verification
    \[\begin{aligned}\frac{d}{dx}\left[e^x(x-1)\right]&=e^x(x-1)+e^x\\ &=xe^x\end{aligned}\]
    📝 The derivative matches the original integrand.
\[\boxed{\int xe^x\,dx=e^x(x-1)+C}\]
❌ Exam Tips and Common Mistakes
  • Remember the minus sign: \(\int u\,dv=uv-\int v\,du\).
  • Differentiate \(u\) and integrate \(dv\) carefully.
  • Do not forget the constant of integration \(C\).
  • Choose \(u\) so that differentiation simplifies the expression whenever possible.
  • If the remaining integral is not simpler, reconsider the choice of \(u\) and \(dv\).
  • For repeated integration by parts, keep track of signs and terms systematically.
🔑 Key Takeaway
∫

Example 15

📋
Table of Contents
6 sections
❓ Question
Evaluate \(\displaystyle\int x\cos x\,dx\)
💡 Concept
🗺️ Roadmap

  1. Identify \(u\) and \(dv\).

  2. Calculate \(du\) and \(v\).

  3. Apply the integration-by-parts formula.

  4. Evaluate the remaining integral and simplify.

  5. Verify the answer by differentiation.

🧩 Solution
  1. Choose \(u\) and \(dv\) Given
    \[I=\int x\cos x\,dx\]
  2. Let
    \[u=x,\quad dv=\cos x\,dx\]
  3. Differentiating \(u\),
    du=dx
  4. Integrating \(dv\),
    \[v=\int\cos x\,dx=\sin x\]
  5. Apply the Integration-by-Parts Formula Using
    \[\int u\,dv=uv-\int v\,du\]
  6. we obtain
    \[\begin{aligned}I&=x\sin x-\int\sin x\,dx\end{aligned}\]
  7. Evaluate the Remaining Integral Since
    \[\int\sin x\,dx=-\cos x+C\]
  8. we have
    \[\begin{aligned}I&=x\sin x-\left(-\cos x\right)+C\\&=x\sin x+\cos x+C\end{aligned}\]
  9. Verification by Differentiation Let
    \[F(x)=x\sin x+\cos x\]
  10. Differentiating using the product rule,
    \[\begin{aligned}F'(x)&=\frac{d}{dx}(x\sin x)+\frac{d}{dx}(\cos x)\\[5pt] &=\sin x+x\cos x-\sin x\\[5pt]&=x\cos x\end{aligned}\]
    📝 This matches the original integrand, confirming the answer.
\[\boxed{\int x\cos x\,dx=x\sin x+\cos x+C}\]
❌ Exam Tips and Common Mistakes
  • Choose \(u=x\) and \(dv=\cos x\,dx\).
  • Remember that \(\int\cos x\,dx=\sin x+C\).
  • The formula contains a minus sign: \(uv-\int v\,du\).
  • Since \(\int\sin x\,dx=-\cos x+C\), subtracting this integral produces \(+\cos x\).
  • Include the constant of integration \(C\).
🔑 Key Takeaway
∫

Exercise 16

📋
Table of Contents
6 sections
❓ Question
Evaluate \(\displaystyle\int \log x\,dx\)
💡 Concept
🗺️ Roadmap

  1. Rewrite the integrand as \(\log x\cdot 1\).

  2. Choose \(u=\log x\) and \(dv=dx\).

  3. Find \(du\) and \(v\).

  4. Apply integration by parts and simplify.

  5. Verify the result by differentiation.

🧩 Solution
  1. Write
    \[I=\int \log x\cdot 1\,dx\]
  2. Let
    \[u=\log x,\qquad dv=dx\]
  3. Differentiating \(u\),
    \[du=\frac1x\,dx\]
  4. Integrating \(dv\),
    \[v=\int dx=x\]
  5. Apply the Integration-by-Parts Formula Using
    \[\int u\,dv=uv-\int v\,du\]
  6. we obtain
    \[\begin{aligned}I&=x\log x-\int x\cdot\frac1x\,dx\\[5pt] &=x\log x-\int 1\,dx\\[5pt] &=x\log x-x+C\end{aligned}\]
    📝 This formula assumes \(x>0\), as is usual when \(\log x\) denotes the natural logarithm.
  7. Verification by Differentiation Let
    \[F(x)=x\log x-x\]
  8. Differentiating,
    \[\begin{aligned}F'(x)&=\frac{d}{dx}(x\log x)-\frac{d}{dx}(x)\\[5pt] &=\log x+x\cdot\frac1x-1\\[5pt] &=\log x+1-1\\[5pt] &=\log x\end{aligned}\]
    📝 This matches the original integrand, confirming the result.
\[\boxed{\int\log x\,dx=x\log x-x+C}\]
⚡ Exam Tips and Common Mistakes
🔑 Key Takeaway
∫

Example 17

📋
Table of Contents
6 sections
❓ Question
Find
\[\int xe^x\,dx\]
💡 Concept
🗺️ Roadmap

  1. Choose \(u=x\) and \(dv=e^x\,dx\).

  2. Calculate \(du\) and \(v\).

  3. Apply the integration-by-parts formula.

  4. Simplify and verify the result by differentiation.

🧩 Solution
  1. Let
    \[I=\int xe^x\,dx\]
  2. Choose the functions. Taking \(x\) as the first function and \(e^x\,dx\) as the second part, we have
    \[u=x,\quad dv=e^x\,dx\]
  3. Find \(du\) and \(v\) Differentiating \(u=x\),
    du=dx
  4. Integrating \(dv=e^x\,dx\),
    \[v=\int e^x\,dx=e^x\]
  5. Apply integration by parts.
    \[ \begin{aligned} I&=uv-\int v\,du\\ &=xe^x-\int e^x\,dx\\ &=xe^x-e^x+C \end{aligned} \]
  6. Take the common factor.
    \[\boxed{\int xe^x\,dx=e^x(x-1)+C}\]
  7. Verification by Differentiation Differentiate the obtained result:
    \[\begin{aligned}\frac{d}{dx}\left[e^x(x-1)+C\right]&=e^x(x-1)+e^x\\ &=xe^x-e^x+e^x\\ &=xe^x\end{aligned}\]
    📝 This matches the original integrand. Hence, the solution is verified.
🔑 Key Takeaway
⚡ Exam Tip
∫

Example 18

📋
Table of Contents
5 sections
❓ Question
Find integral of \(\displaystyle\int \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx\)
💡 Concept
🗺️ Roadmap

  1. Evaluate \(\int \frac{x}{\sqrt{1-x^2}}\,dx\) using substitution.

  2. Apply integration by parts to the original integral.

  3. Simplify the resulting expression.

  4. Verify the answer by differentiation.

🧩 Solution
  1. Evaluate the Auxiliary Integral Consider
    \[J=\int\frac{x}{\sqrt{1-x^2}}\,dx\]
  2. Let
    \[t=1-x^2\]
  3. Differentiating both sides gives
    \[dt=-2x\,dx\]
    \[x\,dx=-\frac12\,dt\]
  4. Substituting into \(J\),
    \[\begin{align}J&=-\frac12\int\frac{1}{\sqrt{t}}\,dt\\[5pt] &=-\frac12\int t^{-1/2}\,dt\\[5pt] &=-\frac12\cdot\frac{t^{1/2}}{1/2}\\[5pt] &=-\sqrt{t}\\[5pt] &=-\sqrt{1-x^2}\tag{1}\end{align}\]
  5. Therefore,
    \[\boxed{J=-\sqrt{1-x^2}+C}\]
  6. Apply Integration by Parts Write the original integral as
    \[I=\int \sin^{-1}x\cdot\frac{x}{\sqrt{1-x^2}}\,dx\]
  7. Choose
    \[u=\sin^{-1}x,\qquad dv=\frac{x}{\sqrt{1-x^2}}\,dx\]
  8. Then
    \[du=\frac{1}{\sqrt{1-x^2}}\,dx\]
  9. From (1),
    \[v=-\sqrt{1-x^2}\]
  10. Using the formula \(\int u\,dv=uv-\int v\,du\), we obtain
    \[ \begin{aligned} I &=uv-\int v\,du\\[5pt] &=-\sqrt{1-x^2}\sin^{-1}x -\int\left(-\sqrt{1-x^2}\right) \frac{1}{\sqrt{1-x^2}}\,dx\\[5pt] &=-\sqrt{1-x^2}\sin^{-1}x+\int 1\,dx\\[5pt] &=-\sqrt{1-x^2}\sin^{-1}x+x+C \end{aligned} \]
  11. Verification by Differentiation Let
    \[F(x)=x-\sqrt{1-x^2}\sin^{-1}x\]
  12. Using the product rule,
    \[\begin{aligned}F'(x)&=1-\frac{d}{dx}\left[\sqrt{1-x^2}\sin^{-1}x\right]\\[5pt] &=1-\left[-\frac{x}{\sqrt{1-x^2}}\sin^{-1}x+\sqrt{1-x^2}\cdot\frac{1}{\sqrt{1-x^2}}\right]\\[5pt] &=1+\frac{x\sin^{-1}x}{\sqrt{1-x^2}}-1\\[5pt] &=\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\end{aligned}\]
    📝 This is the original integrand; hence, the answer is verified
\[\boxed{\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx=x-\sqrt{1-x^2}\sin^{-1}x+C}\]
❌ Common Mistakes
Exam Tips and Key Takeaways
  • Recognise \(\frac{x}{\sqrt{1-x^2}}\) as a suitable part for substitution.
  • Choose \(u=\sin^{-1}x\), because its derivative simplifies the remaining integral.
  • Keep track of the two negative signs when applying integration by parts.
  • For a real-valued integrand, the original expression is defined for \(-1<x<1\).
  • Verify the final result by differentiation whenever possible.
∫

Example 19

📋
Table of Contents
6 sections
❓ Question
Find Integral of \(\displaystyle\int e^x\left(\tan^{-1}x+\frac{1}{1+x^2}\right)\,dx\)
💡 Concept
🗺️ Roadmap

  1. Identify \(f(x)\) from the given integrand.

  2. Differentiate \(f(x)\).

  3. Match \(f(x)+f'(x)\) with the expression inside the bracket.

  4. Apply the integration identity and verify the answer.

🧩 Solution
  1. We are given
    \[I=\int e^x\left(\tan^{-1}x+\frac{1}{1+x^2}\right)\,dx\]
  2. Identify the function. Let
    \[f(x)=\tan^{-1}x\]
  3. Differentiate the function
    \[ f'(x)=\frac{d}{dx}(\tan^{-1}x) =\frac{1}{1+x^2} \]
  4. Match the integrand with the identity Since
    \[f(x)+f'(x)=\tan^{-1}x+\frac{1}{1+x^2}\]
  5. the given integral becomes
    \[I=\int e^x\left[f(x)+f'(x)\right]\,dx\]
  6. Using the identity,
    \[I=e^x f(x)+C\]
  7. Substituting \(f(x)=\tan^{-1}x\), we obtain
    \[ \boxed{ \int e^x\left(\tan^{-1}x+\frac{1}{1+x^2}\right)\,dx =e^x\tan^{-1}x+C } \]
  8. Verification by Differentiation Differentiate the result
    \[\begin{aligned}\frac{d}{dx}\left(e^x\tan^{-1}x+C\right)&=e^x\tan^{-1}x+e^x\frac{1}{1+x^2}\\[5pt] &=e^x\left(\tan^{-1}x+\frac{1}{1+x^2}\right)\end{aligned}\]
    📝 The derivative matches the original integrand. Hence, the answer is verified.
⚡ Exam Tip
🔑 Key Takeaway
∫

Integrals of Some More Types

📋
Table of Contents
5 sections
📖 Introduction
🗂️ Types / Category
  • TYPE 1: INTEGRATION OF \(\sqrt{x^2-a^2}\)
    📌Given\(\displaystyle\int\sqrt{x^2-a^2}\,dx\)
    💡ConceptWe apply integration by parts, taking \(\sqrt{x^2-a^2}\) as the first function and \(1\) as the second function. Recall that
    \[\int u\,dv=uv-\int v\,du\]
    🧩solution
    Apply Integration by Parts
    Let
    \[u=\sqrt{x^2-a^2},\qquad dv=dx\]
    Then
    \[v=x\]
    Also,
    \[\begin{aligned}du&=\frac{d}{dx}\left(\sqrt{x^2-a^2}\right)dx\\[5pt] &=\frac{1}{2\sqrt{x^2-a^2}}\cdot 2x\,dx\\[5pt] &=\frac{x}{\sqrt{x^2-a^2}}\,dx\end{aligned}\]
    Using the integration-by-parts formula,
    \[ \begin{aligned} I&=uv-\int v\,du\\[5pt] &=x\sqrt{x^2-a^2}-\int x\cdot\frac{x}{\sqrt{x^2-a^2}}\,dx\\[5pt] &=x\sqrt{x^2-a^2}-\int\frac{x^2}{\sqrt{x^2-a^2}}\,dx\end{aligned}\]
    Simplify the Integral
    Write \(x^2=(x^2-a^2)+a^2\). Therefore,
    \[\begin{aligned}I&=x\sqrt{x^2-a^2}-\int\frac{(x^2-a^2)+a^2}{\sqrt{x^2-a^2}}\,dx\\[5pt] &=x\sqrt{x^2-a^2}-\int\frac{x^2-a^2}{\sqrt{x^2-a^2}}\,dx-a^2\int\frac{dx}{\sqrt{x^2-a^2}}\\[5pt] &=x\sqrt{x^2-a^2}-\int\sqrt{x^2-a^2}\,dx-a^2\int\frac{dx}{\sqrt{x^2-a^2}}\end{aligned}\]
    Since \(I=\int\sqrt{x^2-a^2}\,dx\), we obtain
    \[2I=x\sqrt{x^2-a^2}-a^2\int\frac{dx}{\sqrt{x^2-a^2}}\]
    Use the Standard Integral
    For \(x>a\), the standard result is
    \[\int\frac{dx}{\sqrt{x^2-a^2}}=\ln\left|x+\sqrt{x^2-a^2}\right|+C\]
    Substituting this result gives
    \[\begin{aligned}2I&=x\sqrt{x^2-a^2}-a^2\ln\left|x+\sqrt{x^2-a^2}\right|+C\end{aligned}\]
    Dividing both sides by \(2\), we obtain
    \[\boxed{\begin{aligned}\int\sqrt{x^2-a^2}\,dx&=\frac{x}{2}\sqrt{x^2-a^2}\\[5pt] &\quad-\frac{a^2}{2}\ln\left|x+\sqrt{x^2-a^2}\right|+C\end{aligned}}\]
    Verification
    Differentiating the result confirms that its derivative is \(\sqrt{x^2-a^2}\). The logarithmic term cancels the additional term arising from differentiating the first term.
  • TYPE 2: INTEGRATION OF \(\sqrt{x^2+a^2}\)
    📌Given
    \[I=\int\sqrt{x^2+a^2}\,dx\]
    🧩Custom
    Apply Integration by Parts
    Take
    \[u=\sqrt{x^2+a^2},\quad dv=dx\]
    Then
    \[v=x,\qquad du=\frac{x}{\sqrt{x^2+a^2}}\,dx\]
    Therefore,
    \[\begin{aligned}I&=x\sqrt{x^2+a^2}-\int\frac{x^2}{\sqrt{x^2+a^2}}\,dx\end{aligned}\]
    Simplify the Integral
    Write \(x^2=(x^2+a^2)-a^2\). Thus,
    \[\begin{aligned}I&=x\sqrt{x^2+a^2}-\int\frac{(x^2+a^2)-a^2}{\sqrt{x^2+a^2}}\,dx\\[5pt] &=x\sqrt{x^2+a^2}-\int\sqrt{x^2+a^2}\,dx+a^2\int\frac{dx}{\sqrt{x^2+a^2}}\\[5pt] &=x\sqrt{x^2+a^2}-I+a^2\int\frac{dx}{\sqrt{x^2+a^2}}\end{aligned}\]
    Rearranging,
    \[2I=x\sqrt{x^2+a^2}+a^2\int\frac{dx}{\sqrt{x^2+a^2}}\]
    Use the Standard Integral
    For \(a>0\),
    \[\int\frac{dx}{\sqrt{x^2+a^2}}=\ln\left|x+\sqrt{x^2+a^2}\right|+C\]
    Substituting,
    \[2I=x\sqrt{x^2+a^2}+a^2\ln\left|x+\sqrt{x^2+a^2}\right|+C\]
    Dividing by \(2\), we get
    \[\boxed{\begin{aligned}\int\sqrt{x^2+a^2}\,dx&=\frac{x}{2}\sqrt{x^2+a^2}\\[5pt] &\quad+\frac{a^2}{2}\ln\left|x+\sqrt{x^2+a^2}\right|+C\end{aligned}}\]
  • TYPE 3: INTEGRATION OF \(\sqrt{a^2-x^2}\)
    📌Given\(\int\sqrt{a^2-x^2}\,dx\)
    🧩Custom
    Apply Integration by Parts
    Take
    \[u=\sqrt{a^2-x^2},\quad dv=dx\]
    Then
    \[v=x\]
    Differentiating \(u\),
    \[\begin{aligned}du&=\frac{1}{2\sqrt{a^2-x^2}}(-2x)\,dx\\[5pt] &=-\frac{x}{\sqrt{a^2-x^2}}\,dx\end{aligned}\]
    Applying integration by parts,
    \[\begin{aligned}I&=x\sqrt{a^2-x^2}-\int x\left(-\frac{x}{\sqrt{a^2-x^2}}\right)dx\\[5pt] &=x\sqrt{a^2-x^2}+\int\frac{x^2}{\sqrt{a^2-x^2}}\,dx\end{aligned}\]
    Simplify the Integral
    Write \(x^2=a^2-(a^2-x^2)\). Then
    \[\begin{aligned}I&=x\sqrt{a^2-x^2}+\int\frac{a^2-(a^2-x^2)}{\sqrt{a^2-x^2}}\,dx\\[5pt] &=x\sqrt{a^2-x^2}+a^2\int\frac{dx}{\sqrt{a^2-x^2}}-\int\sqrt{a^2-x^2}\,dx\\[5pt] &=x\sqrt{a^2-x^2}+a^2\int\frac{dx}{\sqrt{a^2-x^2}}-I\end{aligned}\]
    Rearranging,
    \[2I=x\sqrt{a^2-x^2}+a^2\int\frac{dx}{\sqrt{a^2-x^2}}\]
    Use the Standard Integral
    Recall that
    \[\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)+C,\qquad a>0\]
    Substituting,
    \[2I=x\sqrt{a^2-x^2}+a^2\sin^{-1}\left(\frac{x}{a}\right)+C\]
    Dividing both sides by \(2\), we obtain
    \[\boxed{\begin{aligned}\int\sqrt{a^2-x^2}\,dx&=\frac{x}{2}\sqrt{a^2-x^2}\\[5pt] &\quad+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)+C\end{aligned}}\]
    Verification
    Differentiating the result gives
    \[\begin{aligned}\frac{d}{dx}\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)\right]&=\frac12\sqrt{a^2-x^2}-\frac{x^2}{2\sqrt{a^2-x^2}}\\[5pt] &\quad+\frac{a^2}{2\sqrt{a^2-x^2}}\\[5pt] &=\frac12\sqrt{a^2-x^2}+\frac{a^2-x^2}{2\sqrt{a^2-x^2}}\\[5pt] &=\sqrt{a^2-x^2}\end{aligned}\]
    Hence, the result is verified.
📝 Summary of Standard Results
⚡ Exam Tips and Common Mistakes
🔑 Key Takeaway
∫

Definite Integrals

📋
Table of Contents
5 sections
📘 Definition
✏️ Example
Evaluate
\[\int\limits_1^3 2x\,dx\]
🧩 Solution
  1. Find the antiderivative
    \[F(x)=\int 2x\,dx=x^2\]
  2. Apply the limits.
    \[\begin{aligned}\int\limits_1^3 2x\,dx&=\left[x^2\right]_1^3\\ &=3^2-1^2\\ &=9-1\\ &=8\end{aligned}\]
  3. Therefore,
    \[\boxed{\int\limits_1^3 2x\,dx=8}\]
🏷️ Properties

Important Properties

Important Properties
Numerical Value
A definite integral evaluates to a number, provided the integral exists.
Constant of Integration
The constant of integration \(C\) cancels when evaluating \(F(b)-F(a)\).
Limit Reversal
The order of the limits matters: \(\displaystyle\int\limits_a^b f(x)\,dx=-\int\limits_b^a f(x)\,dx\).
Equal Limits
If the limits are equal, then \(\displaystyle\int\limits_a^a f(x)\,dx=0\), provided the integral is defined.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus allows a definite integral to be evaluated using an antiderivative when its conditions are satisfied.
🔑 Key Takeaway
∫

Fundamental Theorem of Calculus

📋
Table of Contents
9 sections
📖 Area Function
📘 Definition of the Area Function
📌 Understanding the Area Function
✏️ Example
1
Question
Finding an Area Function:
Let \(f(t)=2t\) on \([0,3]\). Find the area function \(A(x)\)
  1. Write the definition
    \[A(x)=\int\limits_0^x 2t\,dt\]
  2. Integrate with respect to \(t\).
    \[ \begin{aligned} A(x)&=\left[t^2\right]_0^x\\ &=x^2-0^2\\ &=x^2 \end{aligned} \]
  3. Hence, the area function is
    \[\boxed{A(x)=x^2,\qquad 0\leq x\leq 3}\]
  4. For example, at \(x=2\), the accumulated area is
    \[A(2)=2^2=4\]
  5. At \(x=3\), the total area is
    \[A(3)=3^2=9\]
🔗 Connection with the Fundamental Theorem of Calculus
The area function provides a connection between integration and differentiation. If \(f\) is continuous on \([a,b]\), then the Fundamental Theorem of Calculus states that the area function
\[A(x)=\int_a^x f(t)\,dt\]
is differentiable for \(a<x<b\), and
\[\boxed{A'(x)=f(x)}\]
This means that the rate of change of the accumulated area at \(x\) equals the value of the function at that point. The result is the first fundamental theorem of calculus.
🗂️ Types / Category
First Fundamental Theorem:
Let \(f\) be a continuous function on the closed interval \([a,b]\), and let \(A(x)\) be its area function defined by

\[A(x)=\int_a^x f(t)\,dt\]
Then \(A(x)\) is differentiable and
\[\boxed{A'(x)=f(x)}\]
for every \(x\) in the interior of \([a,b]\). At the endpoints, the corresponding one-sided derivatives are understood.
Explanation
The area function represents the accumulated signed area between the graph of \(y=f(t)\) and the horizontal axis, from the fixed lower limit \(a\) to the variable upper limit \(x\). Since \(f\) is continuous, the Fundamental Theorem of Calculus tells us that differentiating this accumulated area gives the value of the original function at the upper limit.
\[\frac{d}{dx}\left(\int_a^x f(t)\,dt\right)=f(x)\]
Thus, differentiation reverses the accumulation represented by integration.
Example
Let
\[A(x)=\int_0^x (3t^2+2)\,dt\]
By the first fundamental theorem,
\[\boxed{A'(x)=3x^2+2}\]
This result follows directly without evaluating the integral first.
Second Fundamental Theorem:
Let \(f\) be a continuous function on the closed interval \([a,b]\), and let \(F\) be an antiderivative of \(f\) on that interval. Then
\[F'(x)=f(x)\]
The definite integral of \(f\) from \(a\) to \(b\) is given by
\[\boxed{\int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)}\]
This result is also known as the evaluation theorem for definite integrals.
Explanation
To evaluate a definite integral using this theorem:
  1. Find an antiderivative \(F(x)\) of \(f(x)\).
  2. Substitute the upper limit \(b\) into \(F(x)\) to obtain \(F(b)\).
  3. Substitute the lower limit \(a\) into \(F(x)\) to obtain \(F(a)\).
  4. Subtract the lower-limit value from the upper-limit value.
Example
Evaluate
\[\int_1^3 2x\,dx\]
Solution
Find the antiderivative.
\[F(x)=\int 2x\,dx=x^2\]
Apply the limits.
\[\begin{aligned}\int\limits_1^3 2x\,dx&=\left[x^2\right]_1^3\\ &=F(3)-F(1)\\ &=3^2-1^2\\ &=9-1\\ &=8\end{aligned}\]
Therefore,
\[\boxed{\int_1^3 2x\,dx=8}\]
⚖️ Comparison of the Two Theorems
First Fundamental Theorem Second Fundamental Theorem
Connects the area function with differentiation. Connects definite integration with antiderivatives.
\(\displaystyle \frac{d}{dx}\int_a^x f(t)\,dt=f(x)\) \(\displaystyle \int_a^b f(x)\,dx=F(b)-F(a)\)
Useful for differentiating integrals with variable limits. Useful for calculating the numerical value of a definite integral.
⚡ Exam Tips and Common Mistakes
🔑 Key Takeaway
∫

Example 20

📋
Table of Contents
4 sections
❓ Question
Evaluate
  1. \(\displaystyle\int\limits_2^3 x^2\,dx\)
  2. \(\displaystyle\int\limits_4^9\frac{\sqrt{x}}{(30-x^{3/2})^2}\,dx\)
🧩 Solution
Part (a)

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    Array
    Find the antiderivative.
    \[\int x^2\,dx=\frac{x^3}{3}+C\]

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    Array
\[\boxed{\int_2^3 x^2\,dx=\frac{19}{3}}\]
Part (b)
\(\displaystyle\int\limits_4^9\frac{\sqrt{x}}{(30-x^{3/2})^2}\,dx\)

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    Array
    Choose the substitution
    \[t=30-x^{3/2}\]

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    Array
    Therefore,
    \[\sqrt{x}\,dx=-\frac23\,dt\]

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    Array
    When \(x=9\),
    \[t=30-9^{3/2}=30-27=3\]

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    Array
    Reversing the limits removes the negative sign:
    \[I=\frac23\int_3^{22}t^{-2}\,dt\]

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    Array
    Hence,
    \[\begin{aligned}I&=\frac23\left[-\frac1t\right]_3^{22}\\ &=\frac23\left(-\frac1{22}+\frac13\right)\\ &=\frac23\left(\frac{22-3}{66}\right)\\ &=\frac23\cdot\frac{19}{66}\\ &=\frac{19}{99}\end{aligned}\]
\[\boxed{\int_4^9\frac{\sqrt{x}}{(30-x^{3/2})^2}\,dx=\frac{19}{99}}\]
⚡ Exam Tips and Common Mistakes
🔑 Key Takeaway
∫

Some Properties of Definite Integrals

📋
Table of Contents
4 sections
🗂️ Properties
  • PROPERTY \(P_0\): CHANGE OF THE DUMMY VARIABLE
    📜statement
    \[\boxed{\int_a^b f(x)\,dx=\int_a^b f(t)\,dt}\]
    🧮ProofThe variable of integration is called a dummy variable. Its name does not affect the value of the integral, provided the integrand and limits remain unchanged. For example, if \(F'(x)=f(x)\), then
    \[\begin{aligned}\int_a^b f(x)\,dx&=F(b)-F(a)\\ &=\int_a^b f(t)\,dt\end{aligned}\]
    Therefore,
    \[\boxed{\int_a^b f(x)\,dx=\int_a^b f(t)\,dt}\]
  • PROPERTY \(P_1\): REVERSAL AND EQUAL LIMITS
    📜statement
    \[\boxed{\int_a^b f(x)\,dx=-\int_b^a f(x)\,dx}\]
    also
    \[\boxed{\int_a^a f(x)\,dx=0}\]
    🧮ProofProof of the first result:

    Let \(F'(x)=f(x)\). By the Fundamental Theorem of Integral Calculus,
    \[\begin{aligned}\int\limits_a^b f(x)\,dx&=F(b)-F(a)\\ &=-[F(a)-F(b)]\\&=-\int\limits_b^a f(x)\,dx\end{aligned}\]
    Hence, reversing the limits changes the sign of the integral. Proof of the second result:

    \[\begin{aligned}\int\limits_a^a f(x)\,dx&=F(a)-F(a)\\ &=0\end{aligned}\]
    Thus, an integral with equal upper and lower limits is zero.
  • PROPERTY \(P_2\): SPLITTING AN INTERVAL
    📜statementFor any point \(c\) in the interval, or more generally whenever the integrals are defined,
    \[\boxed{\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx}\]
    🧮ProofLet \(F'(x)=f(x)\). Then
    \[\begin{aligned}\int_a^c f(x)\,dx+\int_c^b f(x)\,dx&=[F(c)-F(a)]+[F(b)-F(c)]\\ &=F(c)-F(a)+F(b)-F(c)\\ &=F(b)-F(a)\\ &=\int_a^b f(x)\,dx\end{aligned}\]
    The terms \(F(c)\) cancel, proving the property.
  • PROPERTY \(P_3\): REPLACING \(x\) BY \(a+b-x\)
    📜statement
    \[\boxed{\int_a^b f(x)\,dx=\int\limits_a^b f(a+b-x)\,dx}\]
    🧮ProofConsider the integral on the right-hand side. Put
    \[t=a+b-x\]
    Then
    \[dt=-dx,\qquad dx=-dt\]
    Changing the limits:
    • When \(x=a\), \(t=b\).
    • When \(x=b\), \(t=a\).
    Therefore,
    \[\begin{aligned}\int\limits_a^b f(a+b-x)\,dx&=\int\limits_b^a f(t)(-dt)\\ &=\int\limits_a^b f(t)\,dt\\ &=\int\limits_a^b f(x)\,dx\end{aligned}\]
    Hence, the property is proved.
  • PROPERTY \(P_4\): REFLECTION ABOUT THE MIDPOINT OF \([0,a]\)
    📜statement
    \[\boxed{\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx}\]
    🧮ProofUsing Property \(P_3\) with the lower limit \(0\) and upper limit \(a\), we obtain
    \[ \begin{aligned} \int\limits_0^a f(x)\,dx &=\int\limits_0^a f(0+a-x)\,dx\\ &=\int\limits_0^a f(a-x)\,dx \end{aligned} \]
    Therefore, the property is proved.
  • PROPERTY \(P_5\): SPLITTING THE INTERVAL \([0,2a]\)
    📜statement
    \[\boxed{\int\limits_0^{2a} f(x)\,dx=\int\limits_0^a f(x)\,dx+\int_0^a f(2a-x)\,dx}\]
    🧮ProofUsing Property \(P_2\), split the integral at \(x=a\):
    \[\int_0^{2a} f(x)\,dx=\int_0^a f(x)\,dx+\int_a^{2a} f(x)\,dx\]
    In the second integral, put
    \[t=2a-x,\qquad dt=-dx\]
    When \(x=a\), \(t=a\); when \(x=2a\), \(t=0\). Hence,
    \[\begin{aligned}\int\limits_a^{2a} f(x)\,dx&=\int\limits_a^0 f(2a-t)(-dt)\\ &=\int\limits_0^a f(2a-t)\,dt\end{aligned}\]
    Renaming the dummy variable \(t\) as \(x\),
    \[\int_a^{2a} f(x)\,dx=\int_0^a f(2a-x)\,dx\]
    Consequently,
    \[\boxed{\int_0^{2a} f(x)\,dx=\int_0^a f(x)\,dx+\int_0^a f(2a-x)\,dx}\]
  • PROPERTY \(P_6\): SPECIAL SYMMETRY ON \([0,2a]\)
    Case (i): If \(f(2a-x)=f(x)\)
    📜statement
    \[\boxed{\int_0^{2a} f(x)\,dx=2\int_0^a f(x)\,dx}\]
    🧮ProofFrom Property \(P_5\),
    \[\int_0^{2a} f(x)\,dx=\int_0^a f(x)\,dx+\int_0^a f(2a-x)\,dx\]
    Since \(f(2a-x)=f(x)\),
    \[\begin{aligned}\int\limits_0^{2a} f(x)\,dx&=\int\limits_0^a f(x)\,dx+\int_0^a f(x)\,dx\\ &=2\int\limits_0^a f(x)\,dx\end{aligned}\]
    Hence, the result is proved.
    Case (ii): If \(f(2a-x)=-f(x)\)
    📜statement
    \[\boxed{\int_0^{2a} f(x)\,dx=0}\]
    🧮ProofAgain, by Property \(P_5\),
    \[\begin{aligned}\int\limits_0^{2a} f(x)\,dx&=\int\limits_0^a f(x)\,dx+\int_0^a f(2a-x)\,dx\\ &=\int\limits_0^a f(x)\,dx-\int_0^a f(x)\,dx\\ &=0\end{aligned}\]
    Thus, the integral is zero when the values of the function cancel in pairs under the reflection \(x\mapsto 2a-x\).
  • PROPERTY \(P_7\): INTEGRALS OF EVEN AND ODD FUNCTIONS
    Let \(f\) be integrable on \([-a,a]\), where \(a>0\).
    (i) Even Function
    A function is called even if
    \[f(-x)=f(x)\]
    📜statement
    \[\boxed{\int_{-a}^{a} f(x)\,dx=2\int_0^a f(x)\,dx}\]
    🧮ProofSplit the integral at zero:
    \[\int_{-a}^{a} f(x)\,dx=\int_{-a}^{0} f(x)\,dx+\int_0^a f(x)\,dx\]
    In the first integral, put \(t=-x\). Then \(dx=-dt\). When \(x=-a\), \(t=a\); when \(x=0\), \(t=0\). Therefore,
    \[\begin{aligned}\int\limits_{-a}^{0} f(x)\,dx&=\int\limits_a^0 f(-t)(-dt)\\ &=\int\limits_0^a f(-t)\,dt\end{aligned}\]
    Since \(f\) is even, \(f(-t)=f(t)\). Thus,
    \[\int\limits_{-a}^{0} f(x)\,dx=\int_0^a f(t)\,dt\]
    Consequently,
    \[\begin{aligned}\int\limits_{-a}^{a} f(x)\,dx&=\int\limits_0^a f(x)\,dx+\int_0^a f(x)\,dx\\ &=2\int\limits_0^a f(x)\,dx\end{aligned}\]
    Hence, the property is proved.
    (ii) Odd Function
    A function is called odd if
    \[f(-x)=-f(x)\]
    📜statement
    \[\boxed{\int_{-a}^{a} f(x)\,dx=0}\]
    🧮ProofAs before, split the integral at zero:
    \[\int_{-a}^{a} f(x)\,dx=\int_{-a}^{0} f(x)\,dx+\int_0^a f(x)\,dx\]
    Using the substitution \(t=-x\) in the first integral,
    \[\begin{aligned}\int\limits_{-a}^{0} f(x)\,dx&=\int\limits_0^a f(-t)\,dt\\ &=-\int\limits_0^a f(t)\,dt\end{aligned}\]
    Therefore,
    \[\begin{aligned}\int\limits_{-a}^{a} f(x)\,dx&=-\int\limits_0^a f(x)\,dx+\int\limits_0^a f(x)\,dx\\ &=0\end{aligned}\]
    Hence, the property is proved.
📝 Summary of Properties
⚡ Exam Tips and Common Mistakes
🔑 Key Takeaway
∫

Example 21

📋
Table of Contents
5 sections
❓ Question
Evaluate \(\displaystyle\int_{-1}^{2}|x^3-x|\,dx\)
💡 Concept
🧩 Solution
  1. Find the Zeros of the Expression We have
    \[x^3-x=x(x^2-1)=x(x-1)(x+1)\]
  2. Setting the expression equal to zero,
    \[x(x-1)(x+1)=0\]
  3. Hence, the zeros are
    \[x=-1,\quad x=0,\quad x=1\]
  4. These points divide the interval \([-1,2]\) into three subintervals:
    \[[-1,0],\quad [0,1],\quad [1,2]\]
  5. Determine the Sign Since \(x^3-x=x(x-1)(x+1)\), its sign on each subinterval is as follows:
    • For \(-1<x<0\), \(x^3-x>0\).
    • For \(0<x<1\), \(x^3-x<0\).
    • For \(1<x\leq2\), \(x^3-x>0\).
  6. Thus,
    \[|x^3-x|=\begin{cases}x^3-x,& -1\leq x\leq0\\ -(x^3-x),& 0\leq x\leq1\\x^3-x,& 1\leq x\leq2\end{cases}\]
  7. Split the Integral Using the sign of the expression on each subinterval,
    \[\begin{aligned}I&=\int_{-1}^{0}(x^3-x)\,dx-\int_0^1(x^3-x)\,dx+\int_1^2(x^3-x)\,dx\end{aligned}\]
  8. Evaluate the First Integral
    \[\begin{aligned}I_1&=\int_{-1}^{0}(x^3-x)\,dx\\ &=\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_{-1}^{0}\\ &=0-\left(\frac14-\frac12\right)\\&=\frac14\end{aligned}\]
  9. Evaluate the Second Integral
    \[\begin{aligned}I_2&=-\int_0^1(x^3-x)\,dx\\ &=-\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_0^1\\ &=-\left(\frac14-\frac12\right)\\ &=\frac14\end{aligned}\]
  10. Evaluate the Third Integral
    \[\begin{aligned}I_3&=\int_1^2(x^3-x)\,dx\\ &=\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_1^2\\ &=\left(4-2\right)-\left(\frac14-\frac12\right)\\ &=2+\frac14\\ &=\frac94\end{aligned}\]
  11. Add the Results
    \[\begin{aligned}I&=I_1+I_2+I_3\\ &=\frac14+\frac14+\frac94\\ &=\frac{11}{4}\end{aligned}\]
\boxed{\int_{-1}^{2}|x^3-x|\,dx=\frac{11}{4}}
⚡ Exam Tip
Exam Tips and Common Mistakes
🔑 Key Takeaway
∫

Example 22

📋
Table of Contents
5 sections
❓ Question
Evaluate \(\displaystyle\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin^2x\,dx\)
💡 Concept
🧩 Solution
Part (a)
  1. Apply the Even-Function Property
    \[\begin{aligned}I&=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin^2x\,dx\\[5pt] &=2\int_0^{\frac{\pi}{4}}\sin^2x\,dx\end{aligned}\]
  2. Use the Double-Angle Identity Recall the identity
    \[\sin^2x=\frac{1-\cos 2x}{2}\]
  3. Substituting this identity,
    \[\begin{aligned}I&=2\int_0^{\frac{\pi}{4}}\frac{1-\cos 2x}{2}\,dx\\[5pt] &=\int_0^{\frac{\pi}{4}}(1-\cos 2x)\,dx\end{aligned}\]
  4. Integrate Integrating each term,
    \[\begin{aligned}I&=\left[x-\frac{\sin 2x}{2}\right]_0^{\frac{\pi}{4}}\\[5pt] &=\left(\frac{\pi}{4}-\frac{\sin\frac{\pi}{2}}{2}\right)-\left(0-\frac{\sin 0}{2}\right)\\[5pt] &=\frac{\pi}{4}-\frac12\end{aligned}\]
  5. Verification Using the Antiderivative Alternatively, use
    \[\int\sin^2x\,dx=\frac{x}{2}-\frac{\sin 2x}{4}+C\]
  6. Applying the limits directly,
    \[\begin{aligned}I&=\left[\frac{x}{2}-\frac{\sin 2x}{4}\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\\[5pt] &=\left(\frac{\pi}{8}-\frac14\right)-\left(-\frac{\pi}{8}+\frac14\right)\\[5pt] &=\frac{\pi}{4}-\frac12\end{aligned}\]
    📝 Both methods give the same result.
\[\boxed{\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin^2x\,dx=\frac{\pi}{4}-\frac12}\]
⚡ Exam Tip
🔑 Key Takeaway
∫

Example 23

📋
Table of Contents
4 sections
❓ Question
Evaluate \(\displaystyle\int\limits_0^\pi\frac{x\sin x}{1+\cos^2x}\,dx\)
💡 Concept
🧩 Solution
  1. Write the Given Integral
    \[I=\int\limits_0^\pi\frac{x\sin x}{1+\cos^2x}\,dx\]
  2. Apply the Substitution \(x=\pi-t\) Using the symmetry property,
    \[I=\int_0^\pi\frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}\,dx\]
  3. Recall that
    \[\sin(\pi-x)=\sin x\]
    and
    \[\cos(\pi-x)=-\cos x\quad\Rightarrow\quad\cos^2(\pi-x)=\cos^2x\]
  4. Therefore,
    \[I=\int\limits_0^\pi\frac{(\pi-x)\sin x}{1+\cos^2x}\,dx\]
  5. Add the Two Expressions We have obtained two expressions for \(I\):
    \[I=\int\limits_0^\pi\frac{x\sin x}{1+\cos^2x}\,dx\]
    \[I=\int\limits_0^\pi\frac{(\pi-x)\sin x}{1+\cos^2x}\,dx\]
  6. Adding these equations,
    \[\begin{aligned}2I&=\int_0^\pi\frac{x\sin x+(\pi-x)\sin x}{1+\cos^2x}\,dx\\[5pt] &=\int_0^\pi\frac{\pi\sin x}{1+\cos^2x}\,dx\\[5pt] &=\pi\int_0^\pi\frac{\sin x}{1+\cos^2x}\,dx\end{aligned}\]
  7. Evaluate the Remaining Integral Let
    \[t=\cos x\]
  8. Then
    \[dt=-\sin x\,dx\]
  9. Changing the limits:
    • When \(x=0\), \(t=1\).
    • When \(x=\pi\), \(t=-1\).
  10. Thus,
    \[ \begin{aligned} \int\limits_0^\pi\frac{\sin x}{1+\cos^2x}\,dx &=-\int\limits_1^{-1}\frac{dt}{1+t^2}\\[5pt] &=\int\limits_{-1}^{1}\frac{dt}{1+t^2}\\[5pt] &=\left[\tan^{-1}t\right]_{-1}^{1}\\[5pt] &=\tan^{-1}(1)-\tan^{-1}(-1)\\[5pt] &=\frac{\pi}{4}+\frac{\pi}{4}\\[5pt] &=\frac{\pi}{2}\end{aligned}\]
  11. Calculate the Final Value Substituting into the expression for \(2I\),
    \[\begin{aligned}2I&=\pi\cdot\frac{\pi}{2}\\[5pt] &=\frac{\pi^2}{2}\end{aligned}\]
  12. Dividing both sides by \(2\),
    \[\boxed{I=\frac{\pi^2}{4}}\]
\[\boxed{\displaystyle\int\limits_0^\pi\frac{x\sin x}{1+\cos^2x}\,dx=\frac{\pi^2}{4}}\]
⚡ Exam Tip
Exam Tips and Key Takeaways
∫

Example 24

📋
Table of Contents
5 sections
❓ Question
Evaluate \(\displaystyle\int\limits_{-1}^{1}\sin^5x\cos^4x\,dx\)
💡 Concept
🧩 Solution
  1. Identify the Integrand Let
    \[f(x)=\sin^5x\cos^4x\]
  2. Recall the trigonometric identities
    \[\sin(-x)=-\sin x\]
    \[\cos(-x)=\cos x\]
  3. Check Whether the Function Is Odd Replacing \(x\) with \(-x\), we obtain
    \[\begin{aligned}f(-x)&=\sin^5(-x)\cos^4(-x)\\ &=(-\sin x)^5(\cos x)^4\\ &=-\sin^5x\cos^4x\\ &=-f(x)\end{aligned}\]
    📝 Therefore, \(f(x)=\sin^5x\cos^4x\) is an odd function.
  4. Apply the Property of Odd Functions The limits of integration are \(-1\) and \(1\), which are symmetric about zero. Hence,
    \[\begin{aligned}I&=\int\limits_{-1}^{1}\sin^5x\cos^4x\,dx\\ &=0\end{aligned}\]
\[\boxed{\displaystyle\int_{-1}^{1}\sin^5x\cos^4x\,dx=0}\]
⚡ Exam Tip
Exam Tips and Common Mistakes
🔑 Key Takeaway
∫

Example 25

📋
Table of Contents
5 sections
❓ Question
Evaluate \(\displaystyle\int\limits_0^{\frac{\pi}{2}}\frac{\sin^4x}{\sin^4x+\cos^4x}\,dx\)
💡 Concept
🧩 Solution

  1. Warning: Array to string conversion in /home/u159659565/domains/academia-aeternum.com/public_html/includes/render_note_article.php on line 6292
    Array
    \[\boxed{\displaystyle\int_0^{\frac{\pi}{2}}\frac{\sin^4x}{\sin^4x+\cos^4x}\,dx=\frac{\pi}{4}}\]
⚡ Exam Tip
Exam Tips and Common Mistakes
🔑 Key Takeaway
·
Integrals — Interactive Learning EngineNCERT Mathematics · Class XII · Chapter 7
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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
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NCERT Class 12 Maths Integrals Notes & Formulas
NCERT Class 12 Maths Integrals Notes & Formulas — Complete Notes & Solutions · academia-aeternum.com
Explore comprehensive NCERT Class 12 Mathematics notes on Integrals, designed to help students understand the fundamental concepts, formulas, properties, and techniques of integration. This chapter covers indefinite and definite integrals, integration by substitution, integration by parts, partial fractions, standard integral formulas, and the Fundamental Theorem of Integral Calculus. Learn how to evaluate definite integrals using important properties, symmetry, even and odd functions, and…
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    integrals — Learning Resources

    Frequently Asked Questions

    Integration is the reverse process of differentiation. It is used to find antiderivatives and evaluate definite integrals, areas under curves, and other mathematical quantities.

    The chapter covers indefinite integrals, definite integrals, integration by substitution, integration by parts, partial fractions, standard integral formulas, properties of definite integrals, and the Fundamental Theorem of Integral Calculus.

    An indefinite integral represents a family of antiderivatives and includes an arbitrary constant C. A definite integral has specified limits and gives a numerical value when the integrand is integrable.

    The integration by parts formula is ?u dv = uv - ?v du. It is useful for integrating products of functions by choosing suitable expressions for u and dv.

    The substitution method simplifies an integral by replacing a suitable expression with a new variable. If t = g(x), then dt = g'(x) dx, which can transform the integral into a simpler form.

    Important standard integrals include ?xn dx = xn?¹/(n+1) + C for n ? -1, ?1/x dx = ln|x| + C, ?e? dx = e? + C, ?sin x dx = -cos x + C, and ?cos x dx = sin x + C.

    The Fundamental Theorem of Integral Calculus connects differentiation and integration. If F'(x) = f(x), then ??? f(x) dx = F(b) - F(a), provided the required continuity and integrability conditions hold.

    Important properties include reversing the limits changes the sign, splitting an interval into smaller intervals, reflecting the variable using x ? a + b - x, and using symmetry for even and odd functions.

    Symmetry can simplify integrals. For an even function, ???? f(x) dx = 2?0? f(x) dx. For an odd function, ???? f(x) dx = 0, provided the function is integrable on the interval.

    These notes help students revise essential formulas, understand integration techniques, practise step-by-step solved examples, and prepare for CBSE board examinations and competitive exams such as JEE Main, JEE Advanced, and CUET.

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