NCERT Class 12 Maths Integrals Exercise 7.8 Solutions

📘 Concept & Theory

Theory and Concept

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This problem is based on the fundamental properties of definite integration, particularly the linearity property of definite integrals and the Fundamental Theorem of Calculus.

Concept 1: Linearity Property of Definite Integrals

The integral of the sum of two functions is equal to the sum of their individual integrals, provided they have the same limits of integration.

\[ \int\limits_a^b [f(x)+g(x)]\,dx = \int\limits_a^b f(x)\,dx+ \int\limits_a^b g(x)\,dx \]

Concept 2: Fundamental Theorem of Calculus

If \(F(x)\) is an antiderivative of \(f(x)\), that is, \(F'(x)=f(x)\), then:

\[ \int\limits_a^b f(x)\,dx=F(b)-F(a) \]

This is also represented by the notation:

\[ \int\limits_a^b f(x)\,dx=[F(x)]_a^b \]

Concept 3: Geometrical Interpretation

A definite integral represents the signed area between the curve and the x-axis over the specified interval. The portion above the x-axis contributes positively, whereas the portion below the x-axis contributes negatively.

In this problem, the integrand is \(y=x+1\), which represents a straight line. It intersects the x-axis at \(x=-1\) and remains above the x-axis throughout the interval \([-1,1]\).

\[\int\limits_{-1}^{1}(x+1)\,dx\]

🗺️ Solution Roadmap

Step-by-step Plan

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Identify the integrand and the limits of integration.
Separate the given expression into two simpler integrals using the linearity property.
Find the antiderivative of each term separately.
Apply the upper and lower limits using the Fundamental Theorem of Calculus.
Simplify the resulting expression to obtain the final answer.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 6 steps

Given —
\[I=\int\limits_{-1}^{1}(x+1)\,dx\]
Apply the linearity property of integration
Splitting the integrand into two separate terms, we get:
\[I=\int\limits_{-1}^{1}x\,dx+\int\limits_{-1}^{1}1\,dx\]
Integrate each term separately
Using the power rule of integration:
\[\int x\,dx=\frac{x^2}{2}+C\]
Also
\[\int 1\,dx=x+C\]
Therefore,
\[I=\left[\frac{x^2}{2}\right]_{-1}^{1}+[x]_{-1}^{1}\]
Apply the limits of integration
Using the property:
\[[F(x)]_a^b=F(b)-F(a)\]
We obtain:
\[I=\left(\frac{1^2}{2}-\frac{(-1)^2}{2}\right)+(1-(-1))\]
Simplify the expression
Since \(1^2=1\) and \((-1)^2=1\):
\[I=\left(\frac{1}{2}-\frac{1}{2}\right)+(1+1)\]

\[I=0+2\]

\[I=2\]

💡 Answer

Final Answer

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\[\boxed{\displaystyle\int\limits_{-1}^{1}(x+1)\,dx=2}\]

🎯 Exam Significance

Exam Significance

›

Fundamental integration properties: This question reinforces the linearity property of definite integrals.
Correct application of limits: It develops accuracy in substituting upper and lower limits and evaluating \(F(b)-F(a)\).
Step-wise presentation: It demonstrates how to split an integrand, integrate individual terms and simplify the result systematically.
Geometrical understanding: It connects definite integration with the area of a geometrical figure, strengthening conceptual understanding.

For JEE Main, JEE Advanced and Other Competitive Examination Aspirants:

Fundamental properties: The linearity property is frequently useful in simplifying more complicated definite integrals.
Geometrical approach: Recognising an integrand as a straight line can reduce an integration problem to a simple area calculation.
Sign and limit accuracy: Substitution of negative lower limits requires careful handling of signs, particularly when evaluating definite integrals.
Problem-solving efficiency: Comparing analytical and geometrical methods helps identify the shortest valid approach in objective examinations.

🚫 Common Mistakes

Common Mistakes

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Incorrect treatment of the lower limit: Always subtract the value at the lower limit from the value at the upper limit.
Sign error: Remember that \(1-(-1)=2\), not zero.
Incorrect interpretation of the integral: A definite integral represents signed area, not necessarily ordinary geometrical area. In this particular problem, the integrand is non-negative throughout the interval.
Confusing an antiderivative with a definite integral: The constant of integration is not required when directly evaluating a definite integral using the Fundamental Theorem of Calculus.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 6 points

The integral of a sum of functions can be evaluated by integrating each function separately.
The Fundamental Theorem of Calculus provides the standard procedure for evaluating definite integrals.
The upper-limit value must be evaluated first, followed by subtraction of the lower-limit value.
A definite integral can often be evaluated geometrically when the integrand represents a simple curve.
The geometrical area under \(y=x+1\) from \(x=-1\) to \(x=1\) is a triangle of area \(2\).
Both analytical integration and geometrical interpretation lead to the same result.

📘 Concept & Theory

Theory and Concept

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This problem is based on the standard integration formula of the reciprocal function and the Fundamental Theorem of Calculus. It also tests the application of logarithmic properties while simplifying the final result.

Concept 1: Standard Integral Formula

The integration of the reciprocal function is given by:

\[ \int\frac{1}{x}\,dx=\ln|x|+C,\quad x\neq 0 \]

Here, \(\ln x\) represents the natural logarithm of \(x\), that is, logarithm to the base \(e\).

Concept 2: Fundamental Theorem of Calculus

If \(F(x)\) is an antiderivative of \(f(x)\), then:

\[ \int\limits_a^b f(x)\,dx=F(b)-F(a) \]

Consequently, for the reciprocal function:

\[ \int\limits_a^b\frac{1}{x}\,dx =\ln|b|-\ln|a| \]

When both limits are positive, this can be simplified using the logarithmic identity:

\[ \ln a-\ln b=\ln\left(\frac{a}{b}\right) \]

Concept 3: Logarithmic Interpretation

The function \(y=\frac{1}{x}\) is positive and decreasing for \(x>0\). Therefore, the definite integral from \(x=2\) to \(x=3\) represents the positive area enclosed by the curve, the x-axis and the vertical lines \(x=2\) and \(x=3\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the integrand and the given limits of integration.
Apply the standard formula for integrating \(1/x\).
Substitute the upper and lower limits using the Fundamental Theorem of Calculus.
Express the difference of logarithms as a single logarithm.
Write the simplified exact value of the definite integral.

✏️ Solution

Complete Solution

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Step-by-step Solution · 5 steps

Given —
\[I=\int\limits_{2}^{3}\frac{1}{x}\,dx\]
Apply the standard integration formula
We know that:
\[\int\frac{1}{x}\,dx=\ln|x|+C\]
Therefore,
\[I=\left[\ln|x|\right]_{2}^{3}\]
Apply the limits of integration
Using the Fundamental Theorem of Calculus:
\[I=\ln|3|-\ln|2|\]
Since both limits are positive, we can remove the modulus signs:
\[I=\ln 3-\ln 2\]
Apply the logarithmic property
Using the identity:
\[\ln a-\ln b=\ln\left(\frac{a}{b}\right)\]
We obtain:
\[I=\ln\left(\frac{3}{2}\right)\]

💡 Answer

Final Answer

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\[ \boxed{\displaystyle\int\limits_{2}^{3}\frac{1}{x}\,dx =\ln\left(\frac{3}{2}\right)} \]

🎯 Exam Significance

Exam Significance

›

Standard integration formula: Reinforces the important formula for integrating the reciprocal function.
Application of definite limits: Develops accuracy in evaluating the antiderivative at the upper and lower limits.
Logarithmic identities: Strengthens the application of logarithmic properties in simplifying mathematical expressions.
Exact mathematical representation: Demonstrates why the logarithmic form is preferred over a rounded decimal approximation.

For JEE Main, JEE Advanced and Other Competitive Examination Aspirants:

Recognition of standard integrals: Identifying \(1/x\) immediately as a logarithmic integral saves calculation time.
Logarithmic simplification: The result can be used in more complex problems involving logarithmic expressions and definite integrals.
Sign analysis: Understanding the positivity of the integrand helps eliminate incorrect answer choices in objective questions.
Conceptual foundation: This integral is a basic building block for problems involving logarithmic functions, properties of definite integrals and area under curves.

🚫 Common Mistakes

Common Mistakes

›

Incorrect integration formula: Remember that the integral of \(1/x\) is \(\ln|x|+C\), not \(x^{-1}/(-1)\).
Reversing the limits: Always evaluate the antiderivative at the upper limit first and subtract its value at the lower limit.
Incorrect logarithmic simplification: Use the identity \(\ln a-\ln b=\ln(a/b)\), not \(\ln(a-b)\).
Ignoring the modulus: The general indefinite integral requires \(\ln|x|\). The modulus can be removed here because both integration limits are positive and the interval does not contain zero.
Confusing logarithm bases: The integration formula gives the natural logarithm, \(\ln x\), not automatically the common logarithm \(\log_{10}x\).

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 5 points

The standard integral of the reciprocal function is \(\int \frac{1}{x}\,dx=\ln|x|+C\).
A definite integral is evaluated by subtracting the antiderivative's value at the lower limit from its value at the upper limit.
The difference of two logarithms can be expressed as the logarithm of their quotient.
The integral is positive because the function \(y=1/x\) remains positive over the given interval.
The exact answer should preferably be retained in logarithmic form rather than replaced by a rounded decimal.

📘 Concept & Theory

Theory and Concept

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This problem is based on the linearity property of definite integration, the power rule of integration and the Fundamental Theorem of Calculus. It involves integrating a polynomial function term by term and evaluating the resulting antiderivative at the specified limits.

Concept 1: Linearity Property of Integration

The integral of a sum or difference of functions can be evaluated by integrating each term separately. Constant coefficients can be taken outside their respective integrals.

\[ \begin{aligned} \int\limits_a^b [af(x)+bg(x)]\,dx ={}&a\int\limits_a^b f(x)\,dx\\ &+b\int\limits_a^b g(x)\,dx \end{aligned} \]

Here, \(a\) and \(b\) represent constants.

Concept 2: Power Rule of Integration

The standard power rule of integration states that:

\[ \int x^n\,dx=\frac{x^{n+1}}{n+1}+C, \quad n\neq -1 \]

Some important applications of this formula are:

\[ \begin{aligned} \int x^3\,dx&=\frac{x^4}{4}+C\\ \int x^2\,dx&=\frac{x^3}{3}+C\\ \int x\,dx&=\frac{x^2}{2}+C\\ \int 1\,dx&=x+C \end{aligned} \]

Concept 3: Fundamental Theorem of Calculus

If \(F(x)\) is an antiderivative of \(f(x)\), then the definite integral is evaluated using:

\[ \int\limits_a^b f(x)\,dx=F(b)-F(a) \]

In other words, the value of the antiderivative at the lower limit is subtracted from its value at the upper limit.

🗺️ Solution Roadmap

Step-by-step Plan

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Identify the polynomial integrand and the given limits \(1\) and \(2\).
Apply the linearity property to separate the polynomial into individual terms.
Integrate each term using the power rule of integration.
Combine the individual antiderivatives to obtain a single antiderivative function.
Substitute the upper limit \(x=2\) and simplify the expression.
Substitute the lower limit \(x=1\) and simplify the expression.
Subtract the lower-limit value from the upper-limit value to obtain the final answer.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 8 steps

Given —
\[I=\int\limits_{1}^{2}(4x^3-5x^2+6x+9)\,dx\]
Apply the linearity property of integration.
Separating the polynomial into individual terms, we get:
\[ \begin{aligned} I={}&4\int\limits_1^2 x^3\,dx -5\int\limits_1^2 x^2\,dx\\ &+6\int\limits_1^2 x\,dx +9\int\limits_1^2 1\,dx \end{aligned} \]
Integrate each term separately
Applying the power rule of integration:
\[ \begin{aligned} I={}&4\left[\frac{x^4}{4}\right]_1^2 -5\left[\frac{x^3}{3}\right]_1^2\\ &+6\left[\frac{x^2}{2}\right]_1^2 +9[x]_1^2 \end{aligned} \]
Simplifying the coefficients:
\[I=\left[x^4-\frac{5}{3}x^3+3x^2+9x\right]_1^2\]
Apply the upper limit \(x=2\)
Substituting \(x=2\) into the antiderivative:
\[ \begin{aligned} F(2) &=(2)^4-\frac{5}{3}(2)^3+3(2)^2+9(2)\\ &=16-\frac{5}{3}\times8+3\times4+18\\ &=16-\frac{40}{3}+12+18\\ &=46-\frac{40}{3}\\ &=\frac{138-40}{3}\\ &=\frac{98}{3} \end{aligned} \]
Therefore, the value of the antiderivative at the upper limit is:
\[F(2)=\frac{98}{3}\]
Apply the lower limit \(x=1\).
Substituting \(x=1\) into the antiderivative:
\[ \begin{aligned} F(1) &=(1)^4-\frac{5}{3}(1)^3+3(1)^2+9(1)\\ &=1-\frac{5}{3}+3+9\\ &=13-\frac{5}{3}\\ &=\frac{39-5}{3}\\ &=\frac{34}{3} \end{aligned} \]
Therefore, the value of the antiderivative at the lower limit is:
\[F(1)=\frac{34}{3}\]
Subtract the lower-limit value from the upper-limit value.
According to the Fundamental Theorem of Calculus:
\[ \begin{aligned} I&=F(2)-F(1)\\ &=\frac{98}{3}-\frac{34}{3}\\ &=\frac{98-34}{3}\\ &=\frac{64}{3} \end{aligned} \]

💡 Answer

Final Answer

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\[\boxed{\displaystyle\int\limits_1^2(4x^3-5x^2+6x+9)\,dx=\frac{64}{3}}\]

🎯 Exam Significance

Exam Significance

›

Polynomial integration: Strengthens the understanding of integrating polynomial functions by applying the power rule.
Integration properties: Demonstrates the application of the linearity property to simplify lengthy expressions.
Fundamental Theorem of Calculus: Provides practice in evaluating antiderivatives at upper and lower limits.
Fractional arithmetic: Develops accuracy in simplifying rational expressions and obtaining exact answers.
Step-wise presentation: Shows a systematic approach that helps students present complete mathematical reasoning in board examinations.

For JEE Main, JEE Advanced and Other Competitive Examination Aspirants:

Integration fundamentals: Polynomial integration is an essential foundation for more advanced definite integral problems.
Calculation accuracy: Handling multiple polynomial terms and fractional coefficients improves algebraic precision.
Antiderivative verification: Differentiating the obtained antiderivative is an effective way to check the correctness of an integration result.
Analytical efficiency: Recognising polynomial terms and applying the power rule directly helps reduce unnecessary calculations.
Conceptual preparation: The same integration principles are used in problems involving area under curves, differential equations and applications of definite integrals.

🚫 Common Mistakes

Common Mistakes

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Incorrect application of the power rule: Increase the exponent by one and divide by the new exponent. For example, \(\int x^3\,dx=x^4/4+C\).
Missing coefficients: Ensure that the constant coefficients \(4\), \(-5\), \(6\) and \(9\) are retained during integration.
Incorrect simplification: Remember that \(6x\) integrates to \(3x^2\), not \(6x^2\).
Incorrect substitution of limits: Substitute the upper and lower limits into the complete antiderivative, not into individual terms inconsistently.
Reversing the order of subtraction: Always calculate \(F(2)-F(1)\), not \(F(1)-F(2)\).
Arithmetic errors: When combining fractions, convert all terms to a common denominator before performing addition or subtraction.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 7 points

The linearity property allows a polynomial integrand to be separated into simpler individual integrals.
The power rule is the fundamental technique for integrating polynomial terms.
Constant terms are integrated using \(\int k\,dx=kx+C\).
The Fundamental Theorem of Calculus requires evaluating the antiderivative at the upper limit and subtracting its value at the lower limit.
Careful handling of coefficients, negative signs and fractions is essential for obtaining accurate results.
Differentiating the antiderivative provides an effective verification method.
The exact value of the given definite integral is \(\frac{64}{3}\).

📘 Concept & Theory

Theory and Concept

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This problem is based on the integration of trigonometric functions, the application of the substitution principle and the Fundamental Theorem of Calculus. It requires integrating a sine function with a linear expression in its argument and evaluating the resulting antiderivative at the given limits.

Concept 1: Standard Integration Formula

The standard integration formula for the sine function is:

\[ \int\sin ax\,dx=-\frac{\cos ax}{a}+C, \quad a\neq 0 \]

In this problem, the argument of the sine function is \(2x\). Therefore, \(a=2\), and the integral becomes:

\[ \int\sin 2x\,dx=-\frac{1}{2}\cos 2x+C \]

Concept 2: Substitution Principle

The substitution method can also be used to evaluate this integral. Let:

\[ t=2x \]

Differentiating both sides:

\[ dt=2\,dx \]

Therefore:

\[ dx=\frac{dt}{2} \]

This explains the appearance of the coefficient \(\frac{1}{2}\) in the antiderivative.

Concept 3: Fundamental Theorem of Calculus

If \(F(x)\) is an antiderivative of \(f(x)\), then:

\[ \int\limits_a^b f(x)\,dx=F(b)-F(a) \]

For the given integral, the antiderivative is evaluated first at \(x=\frac{\pi}{4}\) and then at \(x=0\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the trigonometric function and its argument.
Apply the standard formula for integrating \(\sin ax\).
Obtain the antiderivative, including the coefficient \(1/a\).
Substitute the upper limit \(x=\frac{\pi}{4}\).
Substitute the lower limit \(x=0\).
Use the standard trigonometric values to simplify the expression.
Obtain the final value of the definite integral.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 6 steps

Given —
\[I=\int\limits_0^{\pi/4}\sin 2x\,dx\]
Apply the standard integration formula
We know that:
\[\int\sin ax\,dx=-\frac{1}{a}\cos ax+C\]
Comparing the given integrand \(\sin 2x\) with \(\sin ax\), we have \(a=2\).
Therefore:
\[\int\sin 2x\,dx=-\frac{1}{2}\cos 2x+C\]
Apply the limits of integration
Using the Fundamental Theorem of Calculus:
\[ \begin{aligned} I&=\left[-\frac{1}{2}\cos 2x\right]_0^{\pi/4}\\ &=-\frac{1}{2} \left[\cos 2x\right]_0^{\pi/4} \end{aligned} \]
Substitute the upper limit.
Putting \(x=\frac{\pi}{4}\):
\[ \begin{aligned} \cos 2x &=\cos\left(2\times\frac{\pi}{4}\right)\\ &=\cos\frac{\pi}{2}\\ &=0 \end{aligned} \]
Substitute the lower limit.
Putting \(x=0\):
\[ \begin{aligned} \cos 2x&=\cos(2\times0)\\ &=\cos 0\\ &=1 \end{aligned} \]
Simplify the expression
Substituting both values:
\[ \begin{aligned} I&=-\frac{1}{2} \left[\cos\frac{\pi}{2}-\cos0\right]\\ &=-\frac{1}{2}(0-1)\\ &=-\frac{1}{2}(-1)\\ &=\frac{1}{2} \end{aligned} \]

💡 Answer

Final Answer

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\[\boxed{\displaystyle\int\limits_0^{\pi/4}\sin 2x\,dx=\frac{1}{2}}\]

🎯 Exam Significance

Exam Significance

›

Standard trigonometric integration: Reinforces the formula for integrating \(\sin ax\), an important formula in the Integrals chapter.
Coefficient handling: Develops an understanding of why the coefficient of \(x\) in the argument appears in the denominator of the antiderivative.
Application of limits: Provides practice in evaluating definite integrals involving standard trigonometric angles.
Trigonometric identities and values: Strengthens familiarity with standard values such as \(\cos 0=1\) and \(\cos\frac{\pi}{2}=0\).
Alternative approaches: Demonstrates how substitution can be used to verify an integration result.

For JEE Main, JEE Advanced and Other Competitive Examination Aspirants:

Trigonometric integration: The standard integral of \(\sin ax\) is frequently used in more complex integration problems.
Substitution technique: Changing the variable and transforming the integration limits is an essential problem-solving skill.
Analytical accuracy: Correctly handling coefficients and negative signs prevents common errors in objective questions.
Geometrical understanding: Interpreting definite integrals as areas helps in solving problems involving curves and bounded regions.
Calculation efficiency: Recognising standard angles allows students to simplify expressions without unnecessary calculations.

🚫 Common Mistakes

Common Mistakes

›

Missing the coefficient: The integral of \(\sin 2x\) is \(-\frac{1}{2}\cos 2x+C\), not \(-\cos 2x+C\).
Incorrect sign: Remember that the integral of the sine function introduces a negative sign.
Reversing the limits: Always evaluate the antiderivative at the upper limit first and subtract its value at the lower limit.
Incorrect trigonometric substitution: Remember that \(2\times\frac{\pi}{4}=\frac{\pi}{2}\), not \(\frac{\pi}{4}\).
Forgetting to transform limits: When using substitution, both limits must be converted into the new variable.
Incorrect interpretation of signed area: A definite integral represents signed area. In this problem, the integrand is non-negative throughout the interval, so the integral equals the ordinary geometrical area.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 7 points

The standard integration formula is \(\int\sin ax\,dx=-\frac{1}{a}\cos ax+C\).
The coefficient of the variable inside the trigonometric argument must be accounted for during integration.
The Fundamental Theorem of Calculus is used to evaluate the antiderivative at the given limits.
Standard trigonometric values simplify the evaluation of definite integrals involving multiples of \(\pi\).
The substitution method provides an alternative approach and requires transforming both integration limits.
A definite integral can represent the geometrical area under a trigonometric curve when the function remains non-negative over the interval.
The exact value of the given integral is \(\frac{1}{2}\).

📘 Concept & Theory

Theory and Concept

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This problem is based on the integration of trigonometric functions, the Fundamental Theorem of Calculus and the geometrical interpretation of definite integrals. An important feature of this problem is that the integrand changes sign within the interval of integration. Consequently, the positive and negative contributions cancel each other.

Concept 1: Standard Integration Formula

The standard integration formula for the cosine function is:

\[ \int\cos ax\,dx=\frac{1}{a}\sin ax+C, \quad a\neq 0 \]

For the given function, \(a=2\). Therefore:

\[ \int\cos 2x\,dx=\frac{1}{2}\sin 2x+C \]

Concept 2: Fundamental Theorem of Calculus

If \(F(x)\) is an antiderivative of \(f(x)\), then:

\[ \int\limits_a^b f(x)\,dx=F(b)-F(a) \]

Thus, the definite integral is evaluated by substituting the upper limit first and subtracting the value obtained at the lower limit.

Concept 3: Symmetry and Cancellation of Signed Areas

The function \(y=\cos 2x\) is positive over the interval \(0\leq x<\frac{\pi}{4}\), becomes zero at \(x=\frac{\pi}{4}\), and is negative over \(\frac{\pi}{4}

The positive area above the x-axis and the negative signed area below the x-axis have equal magnitudes. Therefore, their contributions cancel, giving a definite integral equal to zero.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the integrand and the limits of integration.
Apply the standard integration formula for \(\cos ax\).
Obtain the antiderivative, including the coefficient \(\frac{1}{2}\).
Substitute the upper limit \(x=\frac{\pi}{2}\).
Substitute the lower limit \(x=0\).
Use the standard trigonometric values to simplify the expression.
Interpret the result in terms of cancellation of positive and negative signed areas.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 5 steps

Given —
\[I=\int\limits_0^{\pi/2}\cos 2x\,dx\]
Apply the standard integration formula
We know that:
\[We know that:\]
Comparing the given integrand with the standard formula, we have \(a=2\)
Therefore:
\[\int\cos 2x\,dx=\frac{1}{2}\sin 2x+C\]
Apply the limits of integration
Using the Fundamental Theorem of Calculus:
\[ \begin{aligned} I&=\left[\frac{1}{2}\sin 2x\right]_0^{\pi/2}\\ &=\frac{1}{2}\left[\sin 2x\right]_0^{\pi/2} \end{aligned} \]
Substitute the upper limit
Putting \(x=\frac{\pi}{2}\):
\[ \begin{aligned} \sin 2x &=\sin\left(2\times\frac{\pi}{2}\right)\\ &=\sin\pi\\ &=0 \end{aligned} \]
Substitute the lower limit
Putting \(x=0\):
\[ \begin{aligned} \sin 2x&=\sin(2\times0)\\ &=\sin0\\ &=0 \end{aligned} \]
Simplify the expression
Substituting both values:
\[ \begin{aligned} I&=\frac{1}{2}\left[\sin\pi-\sin0\right]\\ &=\frac{1}{2}(0-0)\\ &=\frac{1}{2}\times0\\ &=0 \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^{\pi/2}\cos 2x\,dx=0}\]

🎯 Exam Significance

Exam Significance

›

Trigonometric integration: Reinforces the standard formula for integrating \(\cos ax\).
Application of limits: Develops accuracy in evaluating trigonometric functions at standard angles.
Fundamental Theorem of Calculus: Provides practice in applying upper and lower limits to an antiderivative.
Signed area: Clarifies the difference between the value of a definite integral and the total geometrical area.
Conceptual interpretation: Demonstrates how positive and negative contributions can cancel to produce a zero integral.

For JEE Main, JEE Advanced and Other Competitive Examination Aspirants:

Symmetry properties: Recognising cancellation over symmetric intervals can simplify definite integral problems.
Trigonometric identities: The identity \(\cos(\pi-\theta)=-\cos\theta\) is useful in evaluating integrals through substitution.
Sign analysis: Identifying intervals where an integrand is positive or negative helps avoid confusion between signed and total area.
Multiple approaches: The problem can be solved using direct integration, symmetry or geometrical reasoning.
Objective examination skills: Recognising equal and opposite contributions helps eliminate unnecessary calculations.

🚫 Common Mistakes

Common Mistakes

›

Missing the coefficient: The integral of \(\cos2x\) is \(\frac{1}{2}\sin2x+C\), not \(\sin2x+C\).
Using an incorrect trigonometric value: Remember that \(\sin\pi=0\) and \(\sin0=0\).
Incorrect limit substitution: Substitute the upper limit and lower limit into the complete antiderivative.
Confusing signed area with total area: A zero definite integral does not imply that the region has zero geometrical area.
Ignoring sign changes: The integrand changes from positive to negative at \(x=\frac{\pi}{4}\). This is important when interpreting the result geometrically.
Incorrect symmetry reasoning: Cancellation occurs because the two signed contributions have equal magnitudes and opposite signs, not simply because the interval is symmetric.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

The standard integration formula is \(\int\cos ax\,dx=\frac{1}{a}\sin ax+C\).
The Fundamental Theorem of Calculus gives the value of a definite integral by evaluating the antiderivative at its limits.
The function \(\cos2x\) changes sign at \(x=\frac{\pi}{4}\) within the given interval.
The positive and negative signed areas have equal magnitudes and cancel each other.
A zero definite integral does not mean that the total geometrical area is zero.
The total geometrical area between the curve and the x-axis over the interval is \(1\), whereas the signed integral is \(0\).
Symmetry and substitution provide useful alternative techniques for evaluating definite integrals.

📘 Concept & Theory

Theory and Concept

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This problem is based on the integration of exponential functions and the Fundamental Theorem of Calculus. It also demonstrates the use of algebraic factorisation to express the final answer in a simplified form.

Concept 1: Standard Integration Formula of Exponential Functions

The exponential function \(e^x\) has a unique property: its derivative is equal to the function itself.

\[ \frac{d}{dx}(e^x)=e^x \]

Consequently, its integration is:

\[ \int e^x\,dx=e^x+C \]

More generally, for a constant \(a\neq0\):

\[ \int e^{ax}\,dx=\frac{e^{ax}}{a}+C \]

Concept 2: Fundamental Theorem of Calculus

If \(F(x)\) is an antiderivative of \(f(x)\), then:

\[ \int\limits_a^b f(x)\,dx=F(b)-F(a) \]

Applying this theorem to the exponential function:

\[ \int\limits_a^b e^x\,dx=e^b-e^a \]

Concept 3: Algebraic Factorisation of Exponential Expressions

Exponential expressions can be simplified using the law of exponents:

\[ e^{a+b}=e^a\cdot e^b \]

For example:

\[ e^5=e^{4+1}=e^4\cdot e \]

This property allows us to express \(e^5-e^4\) in a factorised form.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the integrand as the standard exponential function \(e^x\).
Apply the standard integration formula \(\int e^x\,dx=e^x+C\).
Evaluate the antiderivative at the upper limit \(x=5\).
Evaluate the antiderivative at the lower limit \(x=4\).
Subtract the lower-limit value from the upper-limit value.
Apply the laws of exponents to factorise and simplify the final expression.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 7 steps

Given —
\[I=\int\limits_4^5 e^x\,dx\]
Apply the standard integration formula.
We know that:
\[\int e^x\,dx=e^x+C\]
Therefore:
\[I=\left[e^x\right]_4^5\]
Apply the upper limit \(x=5\)
Substituting \(x=5\):
\[F(5)=e^5\]
Apply the lower limit \(x=4\)
Substituting \(x=4\):
\[F(4)=e^4\]
Subtract the lower-limit value from the upper-limit value
Using the Fundamental Theorem of Calculus:
\[ \begin{aligned} I&=F(5)-F(4)\\ &=e^5-e^4 \end{aligned} \]
Factorise the expression
Using the law of exponents:
\[e^5=e^{4+1}=e^4\cdot e\]
Therefore:
\[ \begin{aligned} I&=e^4\cdot e-e^4\\ &=e^4(e-1) \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_4^5 e^x\,dx=e^5-e^4=e^4(e-1)}\]

🎯 Exam Significance

Exam Significance

›

Exponential integration: Reinforces the standard integration formula of the natural exponential function.
Fundamental Theorem of Calculus: Provides practice in evaluating definite integrals using upper and lower limits.
Laws of exponents: Demonstrates how exponential expressions can be factorised into a compact mathematical form.
Exact mathematical answers: Develops the habit of retaining exponential expressions instead of unnecessarily converting them into decimal approximations.
Step-wise presentation: Demonstrates a clear and systematic method of solving definite integration problems.

For JEE Main, JEE Advanced and Other Competitive Examination Aspirants:

Exponential functions: Integration of \(e^x\) is a fundamental component of problems involving exponential and logarithmic functions.
Recognition of standard integrals: Immediate recognition of the antiderivative of \(e^x\) improves calculation speed.
Algebraic simplification: Factorisation using exponential laws is useful in simplifying expressions in objective-type questions.
Growth behaviour: Understanding the increasing nature of exponential functions strengthens the geometrical interpretation of definite integrals.
Foundation for advanced problems: The same integration principle is useful in differential equations, exponential growth models and applications of calculus.

🚫 Common Mistakes

Common Mistakes

›

Incorrect integration formula: Remember that \(\int e^x\,dx=e^x+C\). Unlike \(e^{ax}\), no additional coefficient is required when the exponent is simply \(x\).
Reversing the limits: Always evaluate \(F(5)-F(4)\), not \(F(4)-F(5)\).
Incorrect factorisation: Use \(e^5=e^4\cdot e\), not \(e^4+e\).
Confusing the exponential constant: Remember that \(e\) is a mathematical constant approximately equal to \(2.71828\), not a variable.
Premature decimal approximation: Retain the exact exponential form \(e^4(e-1)\) unless a numerical answer is specifically required.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

The standard integration formula for the natural exponential function is \(\int e^x\,dx=e^x+C\).
The derivative and antiderivative of \(e^x\) are both \(e^x\).
The Fundamental Theorem of Calculus evaluates a definite integral by subtracting the lower-limit value from the upper-limit value.
Exponential expressions can be factorised using the law \(e^{a+b}=e^a\cdot e^b\).
The function \(e^x\) is positive and strictly increasing throughout the real number line.
The definite integral represents the positive area under the exponential curve between the given limits.
The exact value of the integral is \(e^4(e-1)\).

📘 Concept & Theory

Theory and Concept

›

This problem involves the integration of the tangent function. Since tangent can be expressed as the ratio of sine and cosine, we can use substitution to simplify the integral.

Recall the trigonometric identity:

\[ \tan x=\frac{\sin x}{\cos x} \]

The standard integral is:

\[ \int \tan x\,dx=-\ln|\cos x|+C \]

For a definite integral, the Fundamental Theorem of Calculus states that if \(F'(x)=f(x)\), then:

\[ \int\limits_a^b f(x)\,dx=F(b)-F(a) \]

Here, the limits are \(0\) and \(\frac{\pi}{4}\). Since cosine is positive throughout this interval, the absolute value sign can be omitted when evaluating the logarithm.

🗺️ Solution Roadmap

Step-by-step Plan

›

Express \(\tan x\) as \(\frac{\sin x}{\cos x}\).
Apply the substitution \(t=\cos x\).
Use \(dt=-\sin x\,dx\) to obtain the antiderivative.
Apply the upper and lower limits using the Fundamental Theorem of Calculus.
Simplify the logarithmic expression to obtain the final answer.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Given —
\[I=\int\limits_0^{\frac{\pi}{4}}\tan x\,dx\]
Express tangent in terms of sine and cosine
\[I=\int\limits_0^{\frac{\pi}{4}}\frac{\sin x}{\cos x}\,dx\]
Apply substitution
Let
\[t=\cos x\]
Differentiating both sides with respect to \(x\), we get:
\[dt=-\sin x\,dx\]
Therefore,
\[\sin x\,dx=-dt\]
Substitute in the integral
\[ \begin{aligned} I&=-\int\frac{dt}{t}\\ &=-\ln|t|+C\\ &=-\ln|\cos x|+C \end{aligned} \]
Apply the limits of integration
Using the Fundamental Theorem of Calculus:
\[ \begin{aligned} I&=\left[-\ln|\cos x|\right]_0^{\frac{\pi}{4}}\\ &=-\left[\ln\left(\cos\frac{\pi}{4}\right)-\ln(\cos 0)\right] \end{aligned} \]
Substitute the standard trigonometric values.
We know that:
\[\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}\]

\[\cos 0=1\]
Substituting these values:
\[ \begin{aligned} I&=-\left[\ln\left(\frac{1}{\sqrt{2}}\right)-\ln 1\right]\\ &=-\left[\ln\left(\frac{1}{\sqrt{2}}\right)-0\right]\\ &=-\ln\left(\frac{1}{\sqrt{2}}\right) \end{aligned} \]
Simplify using logarithmic properties
\[ \begin{aligned} I&=-\ln(2^{-\frac12})\\ &=-\left(-\frac12\ln 2\right)\\ &=\frac12\ln 2 \end{aligned} \]
Hence, the required value is:
\[\boxed{\displaystyle\int\limits_0^{\frac{\pi}{4}}\tan x\,dx=\frac12\ln 2}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^{\frac{\pi}{4}}\tan x\,dx=\frac12\ln 2}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests the application of substitution, standard trigonometric integrals and the Fundamental Theorem of Calculus.
JEE Main: It reinforces the use of logarithmic integration and the evaluation of definite integrals with standard trigonometric limits.
JEE Advanced: The substitution technique and logarithmic simplification are useful foundations for more involved definite-integral problems.
Conceptual Understanding: It demonstrates how a trigonometric integral can be transformed into a standard logarithmic integral.

🚫 Common Mistakes

Common Mistakes

›

Incorrect substitution: When \(t=\cos x\), remember that \(dt=-\sin x\,dx\). The negative sign must not be omitted.
Incorrect order of limits: Always evaluate the antiderivative at the upper limit first and subtract its value at the lower limit.
Ignoring the logarithmic sign: The expression \(-\ln(1/\sqrt{2})\) is positive, not negative.
Incorrect logarithmic simplification: Use \(\ln(a^n)=n\ln a\) carefully, especially when the exponent is negative.
Omitting the constant during indefinite integration: Include \(+C\) when finding the antiderivative, even though it cancels during definite evaluation.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

The standard integral of tangent is \(\int\tan x\,dx=-\ln|\cos x|+C\).
The substitution \(t=\cos x\) converts the trigonometric integral into the standard form \(\int\frac{dt}{t}\).
The Fundamental Theorem of Calculus requires evaluating the antiderivative at the upper and lower limits.
Logarithmic identities help simplify expressions involving fractional powers.
The final result is positive because the integrand \(\tan x\) is non-negative over the given interval.

📘 Concept & Theory

Theory and Concept

›

This problem involves the integration of the cosecant function. The standard integral of cosecant is obtained by multiplying and dividing the integrand by \(\operatorname{cosec}x-\cot x\), which allows us to express the integral in logarithmic form.

The standard integration formula is:

\[ \int\operatorname{cosec}x\,dx = \ln|\operatorname{cosec}x-\cot x|+C \]

An equivalent form is:

\[ \int\operatorname{cosec}x\,dx = -\ln|\operatorname{cosec}x+\cot x|+C \]

We will use the first form to evaluate the given definite integral.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given integral using the cosecant function.
Multiply and divide the integrand by \(\operatorname{cosec}x-\cot x\).
Use the identity \(\operatorname{cosec}^2x-\cot^2x=1\).
Obtain the logarithmic antiderivative.
Apply the upper and lower limits and simplify using standard trigonometric values.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Given —
\[I=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{4}}\operatorname{cosec}x\,dx\]
Multiply and divide by \(\operatorname{cosec}x-\cot x\).
\[ I= \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\operatorname{cosec}x(\operatorname{cosec}x-\cot x)} {\operatorname{cosec}x-\cot x}\,dx \]
Simplify the numerator using a trigonometric identity
Recall that:
\[\operatorname{cosec}^2x-\cot^2x=1\]
Therefore,
\[ \begin{aligned} I &= \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\operatorname{cosec}^2x-\operatorname{cosec}x\cot x} {\operatorname{cosec}x-\cot x}\,dx\\ &= \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{\operatorname{cosec}x-\cot x} \left(\operatorname{cosec}^2x-\operatorname{cosec}x\cot x\right)\,dx \end{aligned} \]
Identify the derivative of the denominator.
Differentiating \(\operatorname{cosec}x-\cot x\), we get:
\[ \begin{aligned} \frac{d}{dx}(\operatorname{cosec}x-\cot x) &=-\operatorname{cosec}x\cot x+\operatorname{cosec}^2x\\ &=\operatorname{cosec}^2x-\operatorname{cosec}x\cot x \end{aligned} \]
Thus, the integral takes the standard logarithmic form:
\[ I= \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d(\operatorname{cosec}x-\cot x)} {\operatorname{cosec}x-\cot x} \]

\[ I= \left[ \ln|\operatorname{cosec}x-\cot x| \right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} \]
Apply the upper and lower limits
\[ \begin{aligned} I={}& \ln\left|\operatorname{cosec}\frac{\pi}{4} -\cot\frac{\pi}{4}\right|\\ &-\ln\left|\operatorname{cosec}\frac{\pi}{6} -\cot\frac{\pi}{6}\right| \end{aligned} \]
Substitute the standard trigonometric values
We know that:
\[ \operatorname{cosec}\frac{\pi}{4}=\sqrt{2}, \qquad \cot\frac{\pi}{4}=1 \]

\[ \operatorname{cosec}\frac{\pi}{6}=2, \qquad \cot\frac{\pi}{6}=\sqrt{3} \]
Substituting these values:
\[ \begin{aligned} I &=\ln|\sqrt{2}-1|-\ln|2-\sqrt{3}|\\ &=\ln(\sqrt{2}-1)-\ln(2-\sqrt{3}) \end{aligned} \]
Combine the logarithms
Using the logarithmic identity \(\ln a-\ln b=\ln\left(\frac{a}{b}\right)\),
we obtain:
\[ \begin{aligned} I &=\ln\left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)\ &=\ln\left(\frac{(\sqrt{2}-1)(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}\right)\ &=\ln\left((\sqrt{2}-1)(2+\sqrt{3})\right) \end{aligned} \]
Since \((2-\sqrt{3})(2+\sqrt{3})=1\), the final expression is:
\[ \boxed{ I=\ln\left((\sqrt{2}-1)(2+\sqrt{3})\right) } \]

💡 Answer

Final Answer

›

\[\boxed{I=\ln\left((\sqrt{2}-1)(2+\sqrt{3})\right)}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests the standard integration formula for cosecant and the application of the Fundamental Theorem of Calculus.
JEE Main: It develops fluency in logarithmic integration, trigonometric identities and evaluation at standard angles.
JEE Advanced: Recognising the derivative structure in the numerator and denominator is an important technique for solving more complex integrals.
Conceptual Understanding: The solution demonstrates how a suitable algebraic manipulation transforms a trigonometric integral into the standard form \(\int\frac{f'(x)}{f(x)}\,dx\).

🚫 Common Mistakes

Common Mistakes

›

Incorrect standard formula: Remember that the antiderivative is \(\ln|\operatorname{cosec}x-\cot x|+C\), not simply \(\ln|\operatorname{cosec}x|\).
Sign error in differentiation: The derivative of \(\operatorname{cosec}x\) is \(-\operatorname{cosec}x\cot x\), whereas the derivative of \(-\cot x\) is \(+\operatorname{cosec}^2x\).
Incorrect limit substitution: Always subtract the value at the lower limit from the value at the upper limit.
Incorrect trigonometric values: Remember that \(\operatorname{cosec}\frac{\pi}{4}=\sqrt{2}\) and \(\operatorname{cosec}\frac{\pi}{6}=2\).
Incorrect logarithmic simplification: Use \(\ln a-\ln b=\ln(a/b)\) and rationalise the denominator carefully when required.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

The integral of cosecant can be evaluated using a suitable multiplication and division technique.
The identity \(\operatorname{cosec}^2x-\cot^2x=1\) is central to the derivation.
The expression \(\operatorname{cosec}^2x-\operatorname{cosec}x\cot x\) is the derivative of \(\operatorname{cosec}x-\cot x\).
The resulting integral follows the logarithmic pattern \(\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C\).
Careful application of standard trigonometric values and logarithmic properties leads to the final result.

📘 Concept & Theory

Theory and Concept

›

This problem involves the standard integral of an inverse trigonometric function. The integrand has the form of the derivative of the inverse sine function.

The standard differentiation formula is:

\[ \frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}, \qquad -1 < x < 1 \]

Consequently, the standard integration formula is:

\[ \int\frac{dx}{\sqrt{1-x^2}} =\sin^{-1}x+C \]

Here, \(\sin^{-1}x\) represents the inverse sine function, also called arcsine, and not the reciprocal of \(\sin x\).

We can therefore evaluate the given definite integral directly using the Fundamental Theorem of Calculus.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the integrand as the derivative of \(\sin^{-1}x\).
Write the antiderivative using the standard integration formula.
Apply the upper and lower limits of integration.
Substitute the standard inverse trigonometric values.
Simplify to obtain the final answer.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 5 steps

Given —
\[I=\int\limits_0^1\frac{dx}{\sqrt{1-x^2}}\]
Apply the standard integration formula
Using
\[ \int\frac{dx}{\sqrt{1-x^2}} =\sin^{-1}x+C \]
we get:
\[I=\left[\sin^{-1}x\right]_0^1\]
Apply the limits of integration
By the Fundamental Theorem of Calculus:
\[ \begin{aligned} I &=\sin^{-1}(1)-\sin^{-1}(0) \end{aligned} \]
Substitute the standard inverse trigonometric values
We know that:
\[\sin^{-1}(1)=\frac{\pi}{2}\]

\[\sin^{-1}(0)=0\]
Therefore,
\[ \begin{aligned} I &=\frac{\pi}{2}-0\\ &=\frac{\pi}{2} \end{aligned} \]
Hence, the required value is:
\[\boxed{\displaystyle\int\limits_0^1\frac{dx}{\sqrt{1-x^2}}=\frac{\pi}{2}}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\frac{\pi}{2}}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests the direct application of a standard inverse trigonometric integration formula and the Fundamental Theorem of Calculus.
JEE Main: Recognising standard inverse trigonometric derivatives helps solve definite integrals efficiently.
JEE Advanced: The integral provides a useful example of an improper integral with a finite value, since the integrand becomes unbounded at the upper limit.
Conceptual Understanding: It reinforces the relationship between differentiation and integration of inverse trigonometric functions.

🚫 Common Mistakes

Common Mistakes

›

Incorrect standard formula: Do not confuse the integral of \(\frac{1}{\sqrt{1-x^2}}\) with the integral of \(\frac{1}{1+x^2}\).
Misinterpreting inverse sine: The notation \(\sin^{-1}x\) means inverse sine, not \(\frac{1}{\sin x}\).
Incorrect limit substitution: Evaluate the antiderivative at \(x=1\) first and subtract its value at \(x=0\).
Incorrect inverse trigonometric values: Remember that \(\sin^{-1}(1)=\frac{\pi}{2}\) and \(\sin^{-1}(0)=0\).
Ignoring the endpoint behaviour: The integrand is unbounded at \(x=1\), but the improper integral still converges to a finite value.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

The standard integral is \(\int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x+C\).
The Fundamental Theorem of Calculus allows direct evaluation using the antiderivative at the two limits.
The principal value of \(\sin^{-1}(1)\) is \(\frac{\pi}{2}\).
The integrand becomes unbounded at \(x=1\), but the corresponding improper integral converges.
Recognising standard inverse trigonometric forms saves time and reduces unnecessary calculations.

📘 Concept & Theory

Theory and Concept

›

This problem involves the standard integral of an inverse trigonometric function. The integrand is the derivative of the inverse tangent function.

Recall the standard differentiation formula:

\[ \frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2} \]

Therefore, the corresponding standard integration formula is:

\[ \int\frac{dx}{1+x^2}=\tan^{-1}x+C \]

Here, \(\tan^{-1}x\) denotes the inverse tangent function, also called arctangent, and not the reciprocal of \(\tan x\).

We can directly apply this formula and evaluate the definite integral using the Fundamental Theorem of Calculus.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the integrand as the derivative of \(\tan^{-1}x\).
Apply the standard integration formula.
Substitute the upper and lower limits.
Use the standard inverse tangent values.
Simplify to obtain the final result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 5 steps

Given —
\[I=\int\limits_0^1\frac{dx}{1+x^2}\]
Apply the standard integration formula
Using
\[\int\frac{dx}{1+x^2}=\tan^{-1}x+C\]
We obtain:
\[I=\left[\tan^{-1}x\right]_0^1\]
Apply the limits of integration
By the Fundamental Theorem of Calculus:
\[ \begin{aligned} I &=\tan^{-1}(1)-\tan^{-1}(0) \end{aligned} \]
Substitute the standard inverse trigonometric values.
\[\tan^{-1}(1)=\frac{\pi}{4}\]

\[\tan^{-1}(0)=0\]
Substituting these values:
\[ \begin{aligned} I &=\frac{\pi}{4}-0\\ &=\frac{\pi}{4} \end{aligned} \]
Hence, the required value is:
\[\boxed{\displaystyle\int\limits_0^1\frac{dx}{1+x^2}=\frac{\pi}{4}}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\frac{\pi}{4}}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This question tests the direct application of a standard inverse trigonometric integration formula and the Fundamental Theorem of Calculus.
JEE Main: Recognising the standard form \(\frac{1}{1+x^2}\) helps in quickly evaluating definite integrals.
JEE Advanced: This formula is frequently useful in problems involving substitutions, rational functions and inverse trigonometric expressions.
Conceptual Understanding: It strengthens the connection between the derivative of an inverse trigonometric function and its corresponding integral.

🚫 Common Mistakes

Common Mistakes

›

Confusing inverse tangent with reciprocal tangent: The notation \(\tan^{-1}x\) represents the inverse tangent function, not \(\frac{1}{\tan x}\).
Using the wrong standard formula: Remember that the integral of \(\frac{1}{1+x^2}\) is \(\tan^{-1}x+C\), not \(\ln(1+x^2)+C\).
Incorrect limit substitution: Always subtract the value of the antiderivative at the lower limit from its value at the upper limit.
Incorrect inverse trigonometric values: Remember that \(\tan^{-1}(1)=\frac{\pi}{4}\) and \(\tan^{-1}(0)=0\).
Incorrect angle units: In calculus, inverse trigonometric values are conventionally expressed in radians.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

The standard integration formula is \(\int\frac{dx}{1+x^2}=\tan^{-1}x+C\).
The Fundamental Theorem of Calculus converts the definite integral into the difference of antiderivative values at the two limits.
The principal value of \(\tan^{-1}(1)\) is \(\frac{\pi}{4}\).
The integrand is positive and continuous over the given interval, so the definite integral represents a positive finite area.
Recognising standard inverse trigonometric forms makes integration faster and more accurate.

📘 Concept & Theory

Theory and Concept

›

This problem involves the integration of a rational function whose denominator is a difference of two squares. We can evaluate it directly using a standard integration formula.

The standard formula is:

\[ \int\frac{dx}{x^2-a^2} = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C \]

Alternatively, we can derive this result using partial fractions. Recall the algebraic identity:

\[ x^2-a^2=(x-a)(x+a) \]

For the given integral, \(a=1\). Since the integration interval is \([2,3]\), both \(x-1\) and \(x+1\) are positive. Hence, the absolute value signs can be omitted when evaluating the logarithm.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the denominator as a difference of two squares.
Apply the standard integration formula with \(a=1\).
Substitute the upper and lower limits.
Simplify the logarithmic expressions.
Combine the logarithms to obtain the final answer.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 5 steps

Given —
\[I=\int\limits_2^3\frac{dx}{x^2-1}\]
Using
\[\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C\]
Comparing the denominator \(x^2-1\) with \(x^2-a^2\), we get:
\[a^2=1\]

\[a=1\]
Apply the standard integration formula
\[ \begin{aligned} I &=\left[ \frac{1}{2(1)} \ln\left|\frac{x-1}{x+1}\right| \right]_2^3\\ &=\left[ \frac12\ln\left|\frac{x-1}{x+1}\right| \right]_2^3 \end{aligned} \]
Apply the upper and lower limits
\[ \begin{aligned} I &= \frac12\ln\left|\frac{3-1}{3+1}\right| - \frac12\ln\left|\frac{2-1}{2+1}\right|\\ &= \frac12\ln\left(\frac24\right) - \frac12\ln\left(\frac13\right) \end{aligned} \]
Simplify the logarithmic expressions
Hence, the required value is:
\[ \boxed{\displaystyle \int\limits_2^3\frac{dx}{x^2-1} = \frac12\ln\left(\frac32\right)} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\frac12\ln\left(\frac32\right)}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests the application of standard rational integration formulas and the evaluation of definite integrals.
JEE Main: It reinforces the use of logarithmic integration and algebraic identities involving the difference of two squares.
JEE Advanced: The partial fraction method is particularly useful when evaluating more complicated rational functions.
Conceptual Understanding: It demonstrates the connection between standard integration formulas and partial fraction decomposition.

🚫 Common Mistakes

Common Mistakes

›

Missing the factor of one-half: For \(a=1\), the coefficient in the standard formula is \(\frac{1}{2a}=\frac12\), not \(1\).
Incorrect substitution of \(a\): Since \(a^2=1\), take the positive value \(a=1\) in the standard formula.
Omitting absolute value signs: Retain absolute values in the general logarithmic formula. They can be omitted here only after checking that the relevant expressions are positive on the integration interval.
Incorrect order of limits: Always subtract the antiderivative value at \(x=2\) from its value at \(x=3\).
Incorrect logarithmic simplification: Remember that \(\ln a-\ln b=\ln(a/b)\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

The standard integral of \(\frac{1}{x^2-a^2}\) produces a logarithmic expression.
The denominator \(x^2-1\) can be factorised as \((x-1)(x+1)\).
Partial fraction decomposition provides an alternative method for evaluating the integral.
The coefficient \(\frac12\) must be retained throughout the calculation.
Both the standard formula and partial fraction methods lead to the same result.

📘 Concept & Theory

Theory and Concept

›

This problem involves the integration of the square of a trigonometric function. Since the power of cosine is even, we use the power-reduction identity to convert the squared trigonometric function into a simpler expression that can be integrated directly.

The important trigonometric identity is:

\[ \cos^2x=\frac{1+\cos 2x}{2} \]

Substituting this identity transforms the integral into the sum of two elementary integrals.

We also use the standard integration formula:

\[ \int\cos ax\,dx=\frac{\sin ax}{a}+C \]

For a definite integral, the Fundamental Theorem of Calculus states:

\[ \int\limits_a^b f(x)\,dx=F(b)-F(a) \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Apply the power-reduction identity for \(\cos^2x\).
Separate the resulting expression into two integrals.
Integrate each term independently.
Apply the limits \(0\) and \(\frac{\pi}{2}\).
Use the standard sine values to simplify the result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 8 steps

Given —
\[I=\int\limits_0^{\frac{\pi}{2}}\cos^2x\,dx\]
Apply the power-reduction identity
We know that:
\[\cos^2x=\frac{1+\cos 2x}{2}\]
Substituting this identity:
\[ \begin{aligned} I &=\int\limits_0^{\frac{\pi}{2}}\frac{1+\cos 2x}{2}\,dx\\ &=\frac12\int\limits_0^{\frac{\pi}{2}}(1+\cos 2x)\,dx\\ &=\frac12\int\limits_0^{\frac{\pi}{2}}1\,dx +\frac12\int\limits_0^{\frac{\pi}{2}}\cos 2x\,dx \end{aligned} \]
Integrate each term
Using the standard integration formulas:
\[\int 1\,dx=x+C \]

\[\int\cos 2x\,dx=\frac12\sin 2x+C\]
Therefore,
\[\begin{aligned}I&=\left[\frac{x}{2}+\frac14\sin 2x\right]_0^{\frac{\pi}{2}}\end{aligned}\]
Apply the upper and lower limits
\[ \begin{aligned} I &= \left(\frac{\frac{\pi}{2}}{2} +\frac14\sin\left(2\times\frac{\pi}{2}\right)\right)\\ &\quad- \left(\frac02+\frac14\sin(2\times0)\right) \end{aligned} \]
Simplifying the expressions:
\[ I= \frac{\pi}{4}+\frac14\sin\pi -0-\frac14\sin0 \]
Substitute the standard sine values
We know that:
\[\sin\pi=0,\qquad \sin0=0\]
Hence,
\[ \begin{aligned} I &=\frac{\pi}{4}+\frac14(0)-0-\frac14(0)\\ &=\frac{\pi}{4} \end{aligned} \]
Therefore, the required value is:
\[\boxed{\displaystyle\int\limits_0^{\frac{\pi}{2}}\cos^2x\,dx=\frac{\pi}{4}}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\frac{\pi}{4}}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests the application of trigonometric identities and standard integration formulas in evaluating definite integrals.
JEE Main: Power-reduction identities are useful for simplifying integrals involving even powers of sine and cosine.
JEE Advanced: The technique provides a foundation for evaluating more complicated trigonometric integrals involving higher powers and multiple angles.
Conceptual Understanding: It demonstrates how a trigonometric identity can transform a complicated-looking integrand into elementary functions that are easy to integrate.

🚫 Common Mistakes

Common Mistakes

›

Incorrect power-reduction identity: Remember that \(\cos^2x=\frac{1+\cos2x}{2}\), whereas \(\sin^2x=\frac{1-\cos2x}{2}\).
Missing the coefficient: When integrating \(\cos2x\), the result is \(\frac12\sin2x\), not simply \(\sin2x\).
Incorrect distribution of one-half: The factor \(\frac12\) must multiply both terms after applying the identity.
Incorrect limit substitution: Evaluate the antiderivative at the upper limit first and subtract its value at the lower limit.
Incorrect trigonometric values: Remember that \(\sin\pi=0\) and \(\sin0=0\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

The power-reduction identity simplifies the integration of squared trigonometric functions.
The identity \(\cos^2x=\frac{1+\cos2x}{2}\) converts the squared cosine function into a constant and a double-angle cosine term.
The standard formula for integrating \(\cos ax\) is \(\int\cos ax\,dx=\frac{\sin ax}{a}+C\).
The Fundamental Theorem of Calculus is used to evaluate the antiderivative at the specified limits.
The integral is positive because \(\cos^2x\geq0\) throughout the interval.

📘 Concept & Theory

Theory and Concept

›

This problem involves integration by substitution. The numerator contains \(x\), which is proportional to the derivative of the expression \(x^2+1\) in the denominator.

We use the standard logarithmic integration formula:

\[ \int\frac{f'(x)}{f(x)}\,dx =\ln|f(x)|+C \]

In the given integral, let:

\[ f(x)=x^2+1 \]

Differentiating, we obtain:

\[ f'(x)=2x \]

Since the numerator is \(x\), we can introduce a factor of \(\frac12\) and apply the logarithmic integration formula directly.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the expression \(x^2+1\) as the suitable substitution.
Differentiate the substitution to express \(x\,dx\) in terms of \(dt\).
Transform the integral into the standard logarithmic form.
Obtain the antiderivative.
Apply the limits \(2\) and \(3\) and simplify the logarithmic expression.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Given —
\[I=\int\limits_2^3\frac{x}{x^2+1}\,dx\]
Let:
\[t=x^2+1\]
Differentiating both sides with respect to \(x\):
\[\frac{dt}{dx}=2x\]
Therefore,
\[dt=2x\,dx\]

\[x\,dx=\frac{dt}{2}\]
Substitute in the integral.
\[ \begin{aligned} I &=\int\frac{x\,dx}{x^2+1}\\ &=\frac12\int\frac{dt}{t}\\ &=\frac12\ln|t|+C \end{aligned} \]
Substituting \(t=x^2+1\) back into the expression:
\[\int\frac{x\,dx}{x^2+1}=\frac12\ln|x^2+1|+C\]
Since \(x^2+1\) is always positive, the absolute value signs can be omitted:
\[F(x)=\frac12\ln(x^2+1)\]
Apply the limits of integration
\[ \begin{aligned} I &=\left[\frac12\ln(x^2+1)\right]_2^3\\ &=\frac12\ln(3^2+1) -\frac12\ln(2^2+1) \end{aligned} \]
Substitute and simplify
\[ \begin{aligned} I &=\frac12\ln(10)-\frac12\ln(5)\\ &=\frac12[\ln10-\ln5]\\ &=\frac12\ln\left(\frac{10}{5}\right)\\ &=\frac12\ln2 \end{aligned} \]
Hence, the required value is:
\[\boxed{\displaystyle\int\limits_2^3\frac{x}{x^2+1}\,dx=\frac12\ln2}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\frac12\ln2}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests substitution, logarithmic integration and the evaluation of definite integrals.
JEE Main: Recognising the relationship between the numerator and the derivative of the denominator helps solve logarithmic integrals efficiently.
JEE Advanced: The substitution technique is an important foundation for evaluating more complicated rational and algebraic integrals.
Conceptual Understanding: It demonstrates the standard pattern \(\frac{f'(x)}{f(x)}\), which frequently appears in integration problems.

🚫 Common Mistakes

Common Mistakes

›

Missing the factor of one-half: Since \(dt=2x\,dx\), we must write \(x\,dx=\frac{dt}{2}\).
Incorrect substitution: The correct substitution is \(t=x^2+1\), whose derivative is \(2x\).
Forgetting to substitute back: After finding the indefinite integral in terms of \(t\), express the antiderivative in terms of \(x\) before applying the original limits.
Incorrect limit substitution: Evaluate at \(x=3\) first and subtract the value at \(x=2\).
Logarithmic simplification error: Use \(\ln a-\ln b=\ln(a/b)\) to combine the logarithmic terms correctly.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Substitution is effective when the numerator contains the derivative of an expression appearing in the denominator.
The standard logarithmic pattern is \(\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C\).
The factor \(\frac12\) arises because the derivative of \(x^2+1\) is \(2x\), not \(x\).
The expression \(x^2+1\) is always positive, so its logarithm does not require absolute value signs.
Recognising derivative patterns makes definite integration faster and more systematic.

📘 Concept & Theory

Theory and Concept

›

\[ \int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C \]

Concept 2: Inverse tangent integration formula

\[ \int\frac{dx}{x^2+a^2} =\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C \]

In the first part, the numerator is proportional to the derivative of the denominator. In the second part, the denominator can be expressed in the standard form \(x^2+a^2\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Split the numerator into \(2x+3\) and separate the integral into two parts.
Evaluate the first integral using substitution and the logarithmic formula.
Rewrite the denominator of the second integral in the form \(x^2+a^2\).
Apply the standard inverse tangent integration formula.
Combine both antiderivatives and substitute the limits \(0\) and \(1\).
Simplify the resulting logarithmic and inverse trigonometric expressions.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 16 steps

Given —
\[I=\int\limits_0^1\frac{2x+3}{5x^2+1}\,dx\]
Split the integral into two parts.
\[ \begin{aligned} I &=\int\limits_0^1\left(\frac{2x}{5x^2+1} +\frac{3}{5x^2+1}\right)dx\\ &=\int\limits_0^1\frac{2x}{5x^2+1}\,dx +3\int\limits_0^1\frac{dx}{5x^2+1} \end{aligned} \]
Evaluate the first integral
Apply substitution
Let
\[t=5x^2+1\]
Differentiating both sides:
\[dt=10x\,dx\]
Therefore,
\[2x\,dx=\frac{dt}{5}\]
Substitute in the first integral
\[ \begin{aligned} I_1 &=\int\frac{2x}{5x^2+1}\,dx\\ &=\frac15\int\frac{dt}{t}\\ &=\frac15\ln|t|+C\\ &=\frac15\ln|5x^2+1|+C \end{aligned} \]
Since \(5x^2+1\) is always positive, we can omit the absolute value signs:
\[I_1=\frac15\ln(5x^2+1)+C\]
Evaluate the second integral
Rewrite the denominator
\[ \begin{aligned} I_2 &=3\int\frac{dx}{5x^2+1}\\ &=3\int\frac{dx}{5\left(x^2+\frac15\right)}\\ &=\frac35\int\frac{dx}{x^2+\left(\frac1{\sqrt5}\right)^2} \end{aligned} \]
We compare this expression with the standard formula:
\[ \int\frac{dx}{x^2+a^2} =\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C \]
Here,
\[a=\frac1{\sqrt5}\]
Apply the inverse tangent formula
\[ \begin{aligned} I_2 &=\frac35\left[ \sqrt5\tan^{-1}(\sqrt5x) \right]+C\\ &=\frac{3\sqrt5}{5}\tan^{-1}(\sqrt5x)+C\\ &=\frac3{\sqrt5}\tan^{-1}(\sqrt5x)+C \end{aligned} \]
Combining Both Parts
Write the complete antiderivative
\[ \begin{aligned} I ={}&\frac15\ln(5x^2+1)\\ &+\frac3{\sqrt5}\tan^{-1}(\sqrt5x)+C \end{aligned} \]
Apply the limits of integration
\[ \begin{aligned} I =\left[ \frac15\ln(5x^2+1) +\frac3{\sqrt5}\tan^{-1}(\sqrt5x) \right]_0^1 \end{aligned} \]
Substitute the upper limit \(x=1\)
\[ \begin{aligned} F(1) &=\frac15\ln(5(1)^2+1) +\frac3{\sqrt5}\tan^{-1}(\sqrt5)\\ &=\frac15\ln6 +\frac3{\sqrt5}\tan^{-1}(\sqrt5) \end{aligned} \]
Substitute the lower limit \(x=0\).
\[ \begin{aligned} F(0) &=\frac15\ln(5(0)^2+1) +\frac3{\sqrt5}\tan^{-1}(0)\\ &=\frac15\ln1+\frac3{\sqrt5}(0)\\ &=0 \end{aligned} \]
Subtract the lower-limit value from the upper-limit value
\[ \begin{aligned} I &=F(1)-F(0)\\ &=\frac15\ln6 +\frac3{\sqrt5}\tan^{-1}(\sqrt5)-0 \end{aligned} \]
Hence, the required value is:
\[ \boxed{ \displaystyle \frac15\ln6+\frac3{\sqrt5}\tan^{-1}(\sqrt5) } \]

💡 Answer

Final Answer

›

\[ \boxed{ \displaystyle \int\limits_0^1\frac{2x+3}{5x^2+1}\,dx = \frac15\ln6+\frac3{\sqrt5}\tan^{-1}(\sqrt5) } \]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests the ability to split a rational function and apply two different standard integration formulas.
JEE Main: It reinforces substitution, logarithmic integration and inverse trigonometric integration.
JEE Advanced: Recognising multiple integration patterns within a single rational function is useful for solving more challenging problems.
Conceptual Understanding: It demonstrates that a single integral may require different techniques for different terms of the integrand.

🚫 Common Mistakes

Common Mistakes

›

Incorrect substitution factor: From \(t=5x^2+1\), we obtain \(dt=10x\,dx\), so \(2x\,dx=\frac{dt}{5}\).
Missing the factor of \(\frac15\): When taking \(5\) outside the second denominator, the integral becomes \(\frac35\int\frac{dx}{x^2+\frac15}\).
Incorrect inverse tangent coefficient: Since \(a=\frac1{\sqrt5}\), the coefficient becomes \(\frac3{\sqrt5}\), not \(3\sqrt5\).
Incorrect inverse tangent argument: The expression \(\frac{x}{a}\) becomes \(\sqrt5x\), not \(\frac{x}{\sqrt5}\).
Incorrect lower-limit evaluation: Remember that \(\ln1=0\) and \(\tan^{-1}0=0\).
Incorrect order of limits: Always calculate the upper-limit value minus the lower-limit value.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Splitting the numerator can simplify a rational integral into manageable parts.
The logarithmic formula is useful when the numerator is proportional to the derivative of the denominator.
The standard inverse tangent formula applies to expressions of the form \(\frac1{x^2+a^2}\).
When a constant is factored out of the denominator, its reciprocal must be included as a coefficient outside the integral.
Careful tracking of coefficients is essential when combining different integration techniques.

📘 Concept & Theory

Theory and Concept

›

This integral can be evaluated conveniently using the method of integration by substitution. The exponent contains \(x^2\), and the remaining factor \(x\) is proportional to the derivative of \(x^2\).

Substitution principle: If an integral contains a composite function along with the derivative of its inner function, substitution can simplify the expression.

We use the standard exponential integration formula:

\[ \int e^t\,dt=e^t+C \]

The key observation is that the derivative of \(x^2\) is \(2x\), which makes the substitution \(t=x^2\) particularly useful.

🗺️ Solution Roadmap

Step-by-step Plan

›

Choose \(t=x^2\) as the new variable.
Differentiate to obtain \(x\,dx=\frac12dt\).
Transform the original limits into the new variable.
Substitute into the integral and integrate \(e^t\).
Evaluate the resulting expression at the transformed limits.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 8 steps

Given —
\[I=\int\limits_0^1 xe^{x^2}\,dx\]
Apply substitution
Let
\[t=x^2\]
Differentiating both sides with respect to \(x\):
\[dt=2x\,dx\]
Therefore,
\[x\,dx=\frac{dt}{2}\]
Change the limits of integration
When \(x=0\):
\[t=0^2=0\]
When \(x=1\):
\[t=1^2=1\]
Thus, the new limits are \(0\) and \(1\).
Substitute into the integral
\[ \begin{aligned} I &=\int\limits_0^1 e^t\frac{dt}{2}\\ &=\frac12\int\limits_0^1 e^t\,dt \end{aligned} \]
Integrate using the exponential formula
\[ \begin{aligned} I &=\frac12\left[e^t\right]_0^1\\ &=\frac12\left[e^1-e^0\right] \end{aligned} \]
Simplify
Since \(e^1=e\) and \(e^0=1\),
\[\begin{aligned}I&=\frac12(e-1)\end{aligned}\]

💡 Answer

Final Answer

›

\[ \boxed{ \displaystyle \int\limits_0^1 xe^{x^2}\,dx=\frac{e-1}{2} } \]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests the application of substitution in evaluating definite integrals.
JEE Main: Recognising the relationship between the exponent and its derivative helps solve similar exponential integrals quickly.
JEE Advanced: The problem reinforces the importance of transforming integration limits correctly when changing variables.
Conceptual Understanding: It illustrates how a suitable substitution converts a seemingly complicated integrand into a standard exponential integral.

🚫 Common Mistakes

Common Mistakes

›

Missing the factor of 2: From \(t=x^2\), we obtain \(dt=2x\,dx\), so \(x\,dx=\frac12dt\).
Forgetting to change the limits: When \(x\) changes from \(0\) to \(1\), the new variable \(t=x^2\) also changes from \(0\) to \(1\).
Incorrect exponential integration: The integral of \(e^t\) is \(e^t+C\), not \(te^t+C\).
Incorrect evaluation of the lower limit: Remember that \(e^0=1\), not \(0\).
Incorrect final simplification: Subtract the lower-limit value from the upper-limit value to obtain \(e-1\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Use substitution when the derivative of an inner function appears as a factor in the integrand.
For \(t=x^2\), the differential relation is \(x\,dx=\frac12dt\).
In a definite integral, the limits must be transformed whenever the variable of integration is changed.
The exponential function satisfies \(\int e^t\,dt=e^t+C\).
Always evaluate the upper limit first and subtract the lower-limit value.

📘 Concept & Theory

Theory and Concept

›

This problem involves the integration of a rational algebraic function. Since the degree of the numerator is equal to the degree of the denominator, polynomial division is a suitable first step.

After polynomial division, the remaining rational expression can be separated into two parts:

A logarithmic form involving the derivative of the denominator.
A difference-of-squares form that can be integrated using partial fractions or a standard logarithmic formula.

Useful integration formulas:

\[ \int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C \]

\[ \int\frac{du}{u^2-a^2} =\frac1{2a}\ln\left|\frac{u-a}{u+a}\right|+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Perform polynomial division to simplify the rational function.
Express the remaining numerator in terms of the derivative of the denominator and a constant.
Integrate the logarithmic part using substitution.
Complete the square in the quadratic denominator.
Apply the standard formula for the difference of two squares.
Evaluate the antiderivative at the limits \(1\) and \(2\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 20 steps

Given —
\[I=\int\limits_1^2\frac{5x^2}{x^2+4x+3}\,dx\]
Perform polynomial division
Dividing \(5x^2\) by \(x^2+4x+3\), we obtain:
\[\begin{aligned}5x^2&=5(x^2+4x+3)-(20x+15)\end{aligned}\]
Therefore,
\[ \begin{aligned} \frac{5x^2}{x^2+4x+3} &=5-\frac{20x+15}{x^2+4x+3} \end{aligned} \]
Thus,
\[I=\int\limits_1^2\left(5-\frac{20x+15}{x^2+4x+3}\right)dx\]
Express the numerator suitably
The derivative of the denominator \(x^2+4x+3\) is \(2x+4\).
We express the numerator as:
\[\begin{aligned}20x+15&=10(2x+4)-25\end{aligned}\]
Substituting this expression,
\[ \begin{aligned} I &=\int\limits_1^2\left[ 5-\frac{10(2x+4)-25}{x^2+4x+3} \right]dx\\ &=\int\limits_1^2\left[ 5-\frac{10(2x+4)}{x^2+4x+3} +\frac{25}{x^2+4x+3} \right]dx \end{aligned} \]
Separating the integrals:
\[ \begin{aligned} I={}&5\int\limits_1^2dx -10\int\limits_1^2\frac{2x+4}{x^2+4x+3}\,dx\\ &+25\int\limits_1^2\frac{dx}{x^2+4x+3} \end{aligned} \]
Part A: Integrate the logarithmic term
Apply substitution
Let:
\[t=x^2+4x+3\]
Differentiating
\[dt=(2x+4)\,dx\]
Therefore,
\[ \begin{aligned} \int\frac{2x+4}{x^2+4x+3}\,dx &=\int\frac{dt}{t}\\ &=\ln|t|+C\\ &=\ln|x^2+4x+3|+C \end{aligned} \]
For the interval \(1\leq x\leq2\), the denominator is positive, so:
\[\int\frac{2x+4}{x^2+4x+3}\,dx=\ln(x^2+4x+3)+C\]
Part B: Integrate the remaining rational term
Complete the square
\[ \begin{aligned} x^2+4x+3 &=(x+2)^2-4+3\\ &=(x+2)^2-1 \end{aligned} \]
Hence,
\[\int\frac{dx}{x^2+4x+3}=\int\frac{dx}{(x+2)^2-1}\]
Apply the standard formula
Using
\[\int\frac{du}{u^2-a^2}=\frac1{2a}\ln\left|\frac{u-a}{u+a}\right|+C\]
Here, \(u=x+2\) and \(a=1\). Therefore,
\[\begin{aligned}\int\frac{dx}{(x+2)^2-1}&=\frac12\ln\left|\frac{x+2-1}{x+2+1}\right|+C\\ &=\frac12\ln\left|\frac{x+1}{x+3}\right|+C\end{aligned}\]
Combining Both Parts
Write the complete antiderivative
\[ \begin{aligned} I={}&5x-10\ln(x^2+4x+3)\\ &+\frac{25}{2}\ln\left| \frac{x+1}{x+3} \right|+C \end{aligned} \]
Apply the limits of integration
\[\begin{aligned}I=\Bigg[&5x-10\ln(x^2+4x+3)\\ &+\frac{25}{2}\ln\left(\frac{x+1}{x+3}\right)\Bigg]_1^2\end{aligned}\]
Substitute the upper limit \(x=2\).
\[\begin{aligned}F(2)&=10-10\ln(4+8+3)+\frac{25}{2}\ln\left(\frac35\right)\\ &=10-10\ln15+\frac{25}{2}\ln\left(\frac35\right)\end{aligned}\]
Substitute the lower limit \(x=1\)
\[\begin{aligned}F(1)&=5-10\ln(1+4+3)+\frac{25}{2}\ln\left(\frac24\right)\\ &=5-10\ln8+\frac{25}{2}\ln\left(\frac12\right)\end{aligned}\]
Subtract the lower-limit value from the upper-limit value
\[\begin{aligned}I={}&F(2)-F(1)\\ ={}&10-10\ln15+\frac{25}{2}\ln\left(\frac35\right)\\ &-5+10\ln8-\frac{25}{2}\ln\left(\frac12\right)\end{aligned}\]
Combining the constant terms and logarithms
\[\begin{aligned}I&=5+10\ln\left(\frac8{15}\right)+\frac{25}{2}\ln\left(\frac65\right)\end{aligned}\]
Therefore, the required value is:
\[\boxed{\displaystyle5+10\ln\left(\frac8{15}\right)+\frac{25}{2}\ln\left(\frac65\right)}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle5+10\ln\left(\frac8{15}\right)+\frac{25}{2}\ln\left(\frac65\right)}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests polynomial division, algebraic simplification and the application of standard integration formulas.
JEE Main: It reinforces the recognition of logarithmic forms and the integration of rational functions.
JEE Advanced: It develops the ability to combine polynomial division, substitution and partial fraction-related techniques in a single problem.
Conceptual Understanding: The problem demonstrates why algebraic simplification before integration can make a complicated rational expression manageable.

🚫 Common Mistakes

Common Mistakes

›

Incorrect polynomial division: The quotient is \(5\), not \(1\), because the leading terms are \(5x^2\) and \(x^2\).
Incorrect numerator decomposition: The correct identity is \(20x+15=10(2x+4)-25\).
Incorrect completion of the square: Remember that \(x^2+4x+3=(x+2)^2-1\).
Missing the factor of \(\frac12\): The integral of \(\frac1{u^2-1}\) contains the coefficient \(\frac12\) in its logarithmic result.
Confusing an algebraic reciprocal with its integral: The integral of \(\frac1{u^2-1}\) is logarithmic; it is not simply \(\frac1{u^2-1}\).
Incorrect evaluation of limits: Always calculate \(F(2)-F(1)\), preserving the signs of every term.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

When the numerator and denominator have the same degree, polynomial division should be considered first.
Expressing the numerator in terms of the derivative of the denominator helps identify logarithmic integration.
Completing the square can transform a quadratic expression into a standard difference-of-squares form.
The integral of \(\frac1{u^2-a^2}\) produces a logarithmic expression with a coefficient of \(\frac1{2a}\).
Careful algebraic simplification is essential for obtaining the correct definite integral.

📘 Concept & Theory

Theory and Concept

›

This problem involves the integration of a sum of three elementary functions. We apply the linearity property of integration and integrate each term separately.

Concept 1: Linearity of integration

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

Concept 2: Standard integration formulas

The following formulas are used:

\[ \int\sec^2x\,dx=\tan x+C \]

\[ \int x^n\,dx=\frac{x^{n+1}}{n+1}+C \quad (n\neq-1) \]

\[ \int k\,dx=kx+C \]

By applying these formulas independently, we can obtain the antiderivative and then evaluate it at the given limits.

🗺️ Solution Roadmap

Step-by-step Plan

›

Separate the given integrand into three individual terms.
Integrate \(2\sec^2x\) using the standard secant-squared formula.
Integrate \(x^3\) using the power rule.
Integrate the constant term \(2\).
Combine the antiderivatives.
Apply the upper and lower limits, \(x=\frac{\pi}{4}\) and \(x=0\).
Simplify the resulting expression.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Given —
\[I=\int\limits_0^{\frac{\pi}{4}}\left(2\sec^2x+x^3+2\right)dx\]
Separate the integral
Using the linearity property
\[ \begin{aligned} I={}&2\int\limits_0^{\frac{\pi}{4}}\sec^2x\,dx\\ &+\int\limits_0^{\frac{\pi}{4}}x^3\,dx +2\int\limits_0^{\frac{\pi}{4}}dx \end{aligned} \]
Part A: Integration of the first term
Integrate \(2\sec^2x\).
Using the standard formula:
\[\int\sec^2x\,dx=\tan x+C\]
We obtain:
\[\begin{aligned}2\int\sec^2x\,dx&=2\tan x\end{aligned}\]
Part B: Integration of the second term
Integrate \(x^3\).
Using the power rule:
\[\begin{aligned}\int x^3\,dx&=\frac{x^{3+1}}{3+1}\\ &=\frac{x^4}{4}\end{aligned}\]
Part C: Integration of the third term
Integrate the constant term
\[\begin{aligned}\int 2\,dx=2x\end{aligned}\]
Combine all three antiderivatives
\[\begin{aligned}\int(2\sec^2x+x^3+2)\,dx&=2\tan x+\frac{x^4}{4}+2x+C\end{aligned}\]
Apply the limits of integration
\[I=\left[2\tan x+\frac{x^4}{4}+2x\right]_0^{\frac{\pi}{4}}\]
Substitute the upper limit \(x=\frac{\pi}{4}\)
\[\begin{aligned}F\left(\frac{\pi}{4}\right)&=2\tan\frac{\pi}{4}+\frac{\left(\frac{\pi}{4}\right)^4}{4}+2\left(\frac{\pi}{4}\right)\\ &=2(1)+\frac{\pi^4}{4^4\times4}+\frac{\pi}{2}\\ &=2+\frac{\pi^4}{1024}+\frac{\pi}{2}\end{aligned}\]
Here, we have used:
\[\tan\frac{\pi}{4}=1\]
Substitute the lower limit \(x=0\).
\[\begin{aligned}F(0)&=2\tan0+\frac{0^4}{4}+2(0)\\ &=0+0+0\\ &=0\end{aligned}\]
Subtract the lower-limit value from the upper-limit value.
\[\begin{aligned}I&=F\left(\frac{\pi}{4}\right)-F(0)\\ &=2+\frac{\pi^4}{1024}+\frac{\pi}{2}-0\end{aligned}\]
Hence, the required value is:
\[\boxed{\displaystyle2+\frac{\pi}{2}+\frac{\pi^4}{1024}}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^{\frac{\pi}{4}}(2\sec^2x+x^3+2)\,dx=2+\frac{\pi}{2}+\frac{\pi^4}{1024}}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests the linearity property of integration and the application of standard integration formulas.
JEE Main: It reinforces the quick recognition of standard integrals involving trigonometric functions, polynomial expressions and constants.
JEE Advanced: It strengthens algebraic accuracy and the systematic evaluation of definite integrals involving multiple function types.
Conceptual Understanding: It demonstrates that a sum of elementary functions can be integrated term by term without requiring substitution or integration by parts.

🚫 Common Mistakes

Common Mistakes

›

Incorrect trigonometric integration: Remember that \(\int\sec^2x\,dx=\tan x+C\).
Incorrect application of the power rule: The integral of \(x^3\) is \(\frac{x^4}{4}\), not \(\frac{x^3}{3}\).
Missing the coefficient: The constant multiplier \(2\) must be retained when integrating \(2\sec^2x\) and \(2\).
Incorrect fourth-power calculation: The expression \(\frac14\left(\frac{\pi}{4}\right)^4\) simplifies to \(\frac{\pi^4}{1024}\).
Confusing tangent and inverse tangent: The lower-limit evaluation uses \(\tan0=0\), not \(\tan^{-1}0\).
Incorrect final simplification: Retain the fourth power of \(\pi\) in the polynomial term; do not reduce \(\pi^4\) to \(\pi^2\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

The linearity property allows a sum of functions to be integrated term by term.
The integral of \(\sec^2x\) is \(\tan x+C\).
The power rule is applicable to polynomial terms.
Constant terms are integrated by multiplying the constant by the variable.
Careful substitution of fractional limits is essential to avoid errors involving powers and coefficients.

📘 Concept & Theory

Theory and Concept

›

This problem involves the application of trigonometric identities to simplify a definite integral. The integrand contains the squares of sine and cosine functions with the same half-angle. Using a double-angle identity, we can convert the expression into a simple cosine function.

Important trigonometric identity:

\[ \cos 2\theta=\cos^2\theta-\sin^2\theta \]

Rearranging:

\[ \sin^2\theta-\cos^2\theta=-\cos2\theta \]

Putting \(\theta=\frac{x}{2}\), we obtain:

\[ \sin^2\frac{x}{2}-\cos^2\frac{x}{2} =-\cos x \]

We also use the standard integration formula:

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

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the double-angle trigonometric identity.
Substitute \(\theta=\frac{x}{2}\) into the identity.
Replace the original integrand with \(-\cos x\).
Integrate using the standard cosine integration formula.
Evaluate the antiderivative at \(x=0\) and \(x=\pi\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Given —
\[I=\int\limits_0^\pi\left(\sin^2\frac{x}{2}-\cos^2\frac{x}{2}\right)dx\]
Apply the double-angle identity
We know that:
\[\cos2\theta=\cos^2\theta-\sin^2\theta\]
Therefore,
\[\sin^2\theta-\cos^2\theta=-\cos2\theta\]
Putting \(\theta=\frac{x}{2}\):
\[ \begin{aligned} \sin^2\frac{x}{2}-\cos^2\frac{x}{2} &=-\cos\left(2\times\frac{x}{2}\right)\\ &=-\cos x \end{aligned} \]
Substitute the simplified expression
\[I=-\int\limits_0^\pi\cos x\,dx\]
Integrate:
Using:
\[\int\cos x\,dx=\sin x+C\]
We obtain:
\[I=-\left[\sin x\right]_0^\pi\]
Apply the upper limit \(x=\pi\).
\[\sin\pi=0\]
Apply the lower limit \(x=0\).
\[\sin0=0\]
Calculate the definite integral
\[\begin{aligned}I&=-[\sin\pi-\sin0]\\ &=-[0-0]\\ &=0\end{aligned}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^\pi\left(\sin^2\frac{x}{2}-\cos^2\frac{x}{2}\right)dx=0}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests the application of double-angle identities in evaluating definite integrals.
JEE Main: It develops the ability to simplify trigonometric expressions before integration.
JEE Advanced: It illustrates how trigonometric identities and symmetry of signed areas can simplify definite integral problems.
Conceptual Understanding: It demonstrates that a definite integral can be zero even when the integrand is not identically zero, because positive and negative signed areas may cancel.

🚫 Common Mistakes

Common Mistakes

›

Incorrect identity: Remember that \(\cos2\theta=\cos^2\theta-\sin^2\theta\).
Missing the negative sign: The expression \(\sin^2\theta-\cos^2\theta\) equals \(-\cos2\theta\), not \(+\cos2\theta\).
Incorrect angle substitution: When \(\theta=\frac{x}{2}\), the double angle \(2\theta\) becomes \(x\).
Incorrect integration: The integral of \(\cos x\) is \(\sin x+C\).
Incorrect limit evaluation: Both \(\sin\pi\) and \(\sin0\) are zero.
Confusing signed area with total area: A zero definite integral indicates cancellation of positive and negative signed areas, not necessarily an absence of geometric area.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Double-angle identities can simplify trigonometric expressions before integration.
The expression \(\sin^2\frac{x}{2}-\cos^2\frac{x}{2}\) simplifies to \(-\cos x\).
The definite integral is evaluated by subtracting the antiderivative value at the lower limit from its value at the upper limit.
A definite integral can have a zero value because positive and negative signed areas cancel each other.
Recognising suitable trigonometric identities can significantly reduce the steps required to solve an integral.

📘 Concept & Theory

Theory and Concept

›

This problem involves the integration of a rational function that can be separated into two simpler integrals. The first integral follows the logarithmic integration formula because the numerator is proportional to the derivative of the denominator. The second integral follows the standard inverse tangent formula.

Concept 1: Logarithmic integration

\[ \int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C \]

Concept 2: Inverse tangent integration

\[ \int\frac{dx}{x^2+a^2} =\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C \]

By identifying these two standard forms, the given integral can be evaluated without using partial fractions.

🗺️ Solution Roadmap

Step-by-step Plan

›

Split the numerator into \(6x\) and \(3\).
Evaluate the first integral using the substitution \(t=x^2+4\).
Evaluate the second integral using the inverse tangent formula.
Combine both antiderivatives.
Apply the limits \(0\) and \(2\).
Simplify the logarithmic and inverse trigonometric expressions.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 15 steps

Given —
\[I=\int\limits_0^2\frac{6x+3}{x^2+4}\,dx\]
Separate the integral
\[\begin{aligned}I&=\int\limits_0^2\left(\frac{6x}{x^2+4}+\frac3{x^2+4}\right)dx\\ &=\int\limits_0^2\frac{6x}{x^2+4}\,dx+3\int\limits_0^2\frac{dx}{x^2+4}\end{aligned}\]
Part A: Evaluate the logarithmic integral
Apply substitution
Let
\[t=x^2+4\]
Differentiating both sides:
\[dt=2x\,dx\]
Therefore,
\[6x\,dx=3\,dt\]
Substitute into the first integral
\[\begin{aligned}I_1&=\int\frac{6x}{x^2+4}\,dx\\ &=3\int\frac{dt}{t}\\ &=3\ln|t|+C\\ &=3\ln|x^2+4|+C\end{aligned}\]
Since \(x^2+4\) is always positive, the absolute value signs can be omitted:
\[I_1=3\ln(x^2+4)+C\]
Part B: Evaluate the inverse tangent integral
Apply the standard formula
\[I_2=3\int\frac{dx}{x^2+4}\]
Comparing the denominator with \(x^2+a^2\), we have:
\[a^2=4\implies a=2\]
Using the standard formula
\[\begin{aligned}I_2&=3\left[\frac12\tan^{-1}\left(\frac{x}{2}\right)\right]+C\\ &=\frac32\tan^{-1}\left(\frac{x}{2}\right)+C\end{aligned}\]
Combining Both Parts
Write the complete antiderivative
\[ \begin{aligned} I={}&3\ln(x^2+4)\\ &+\frac32\tan^{-1}\left(\frac{x}{2}\right)+C \end{aligned} \]
Apply the limits of integration
\[I=\left[3\ln(x^2+4)+\frac32\tan^{-1}\left(\frac{x}{2}\right)\right]_0^2\]
Substitute the upper limit \(x=2\).
\[\begin{aligned}F(2)&=3\ln(2^2+4)+\frac32\tan^{-1}\left(\frac22\right)\\ &=3\ln8+\frac32\tan^{-1}(1)\\ &=3\ln8+\frac32\cdot\frac{\pi}{4}\\&=3\ln8+\frac{3\pi}{8}\end{aligned}\]
Here, we have used:
\[\tan^{-1}(1)=\frac{\pi}{4}\]
Substitute the lower limit \(x=0\).
\[\begin{aligned}F(0)&=3\ln(0^2+4)+\frac32\tan^{-1}\left(\frac02\right)\\ &=3\ln4+\frac32\tan^{-1}(0)\\ &=3\ln4+0\end{aligned}\]
Subtract the lower-limit value from the upper-limit value
\[\begin{aligned}I&=F(2)-F(0)\\ &=3\ln8+\frac{3\pi}{8}-3\ln4\\ &=3(\ln8-\ln4)+\frac{3\pi}{8}\\ &=3\ln\left(\frac84\right)+\frac{3\pi}{8}\\ &=3\ln2+\frac{3\pi}{8}\end{aligned}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^2\frac{6x+3}{x^2+4}\,dx=3\ln2+\frac{3\pi}{8}}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests the ability to split a rational function and apply logarithmic and inverse tangent integration formulas.
JEE Main: It reinforces substitution and the recognition of standard inverse trigonometric integration forms.
JEE Advanced: It develops accuracy in combining different integration techniques within a single definite integral.
Conceptual Understanding: It demonstrates how identifying the derivative of the denominator can simplify rational integration.

🚫 Common Mistakes

Common Mistakes

›

Incorrect substitution: From \(t=x^2+4\), we obtain \(dt=2x\,dx\), so \(6x\,dx=3\,dt\).
Missing the coefficient: The factor \(3\) must be retained when evaluating the second integral.
Incorrect inverse tangent formula: For \(x^2+4\), the value of \(a\) is \(2\), not \(4\).
Incorrect coefficient: The second antiderivative is \(\frac32\tan^{-1}(x/2)\), not \(\frac12\tan^{-1}(x/2)\).
Incorrect logarithmic simplification: Use \(\ln8-\ln4=\ln2\).
Incorrect limit evaluation: The lower-limit contribution includes \(3\ln4\), which must be subtracted from the upper-limit expression.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Splitting the numerator can reveal two standard integration forms.
Substitution is effective when the numerator contains the derivative of the denominator.
The integral of \(\frac1{x^2+a^2}\) produces an inverse tangent expression.
Logarithmic properties help simplify the final result.
In definite integration, every term of the antiderivative must be evaluated at both limits.

📘 Concept & Theory

Theory and Concept

›

This problem involves the integration of two different types of functions. The integral is evaluated by splitting it into two parts and applying the appropriate integration technique to each part.

Concept 1: Integration by Parts

Integration by parts is based on the product rule of differentiation and is expressed as:

\[ \int u\,dv=uv-\int v\,du \]

It is particularly useful when the integrand is a product of two functions, such as \(xe^x\).

Concept 2: Integration of a Sine Function

The standard formula for integrating a sine function is:

\[ \int\sin(ax)\,dx=-\frac{\cos(ax)}{a}+C \]

Here, the coefficient of \(x\) inside the sine function is \(\frac{\pi}{4}\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Separate the given integral into exponential and trigonometric parts.
Apply integration by parts to \(\int xe^x\,dx\).
Integrate \(\sin\frac{\pi x}{4}\) using the standard formula.
Combine the two antiderivatives.
Evaluate the resulting expression at the limits \(0\) and \(1\).
Simplify the final expression.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 13 steps

Given —
\[I=\int\limits_0^1\left(xe^x+\sin\frac{\pi x}{4}\right)dx\]
Split the integral
\[I=\int\limits_0^1xe^x\,dx+\int\limits_0^1\sin\frac{\pi x}{4}\,dx\]
Let:
\[I_1=\int\limits_0^1xe^x\,dx\]
and
\[I_2=\int\limits_0^1\sin\frac{\pi x}{4}\,dx\]
Part A: Integration of \(xe^x\)
Apply integration by parts
We use the formula:
\[\int u\,dv=uv-\int v\,du\]
Choose:
\[u=x,\quad dv=e^x\,dx\]
Differentiating and integrating respectively:
\[du=dx,\qquad v=e^x\]
Substitute into the integration by parts formula
\[\begin{aligned}I_1&=\int xe^x\,dx\\ &=xe^x-\int e^x\,dx\\ &=xe^x-e^x+C\\ &=e^x(x-1)+C\end{aligned}\]
Evaluate the first integral between 0 and 1
\[\begin{aligned}I_1&=\left[e^x(x-1)\right]_0^1\\ &=\left[e^1(1-1)\right]-\left[e^0(0-1)\right]\\ &=e(0)-1(-1)\\ &=0+1\\ &=1\end{aligned}\]
Part B: Integration of the Sine Function
Evaluate the second integral
\[I_2=\int\limits_0^1\sin\frac{\pi x}{4}\,dx\]
Using the standard formula:
\[\int\sin(ax)\,dx=-\frac{\cos(ax)}{a}+C\]
Here:
\[a=\frac{\pi}{4}\]
Therefore,
\[\begin{aligned}I_2&=-\frac{1}{\pi/4}\cos\frac{\pi x}{4}\Bigg|_0^1\\ &=-\frac4\pi\left[\cos\frac{\pi x}{4}\right]_0^1\\ &=-\frac4\pi\left[\cos\frac{\pi}{4}-\cos0\right]\end{aligned}\]
Substitute the standard trigonometric values
We know that:
\[\cos\frac{\pi}{4}=\frac{\sqrt2}{2},\qquad \cos0=1\]
Substituting these values:
\[\begin{aligned}I_2&=-\frac4\pi\left(\frac{\sqrt2}{2}-1\right)\\ &=\frac4\pi\left(1-\frac{\sqrt2}{2}\right)\end{aligned}\]
Combining Both Parts
Add the values of the two integrals
\[I=I_1+I_2\]

\[\begin{aligned}I&=1+\frac4\pi\left(1-\frac{\sqrt2}{2}\right)\end{aligned}\]
Thus, the required definite integral is:
\[\boxed{I=1+\frac4\pi\left(1-\frac{\sqrt2}{2}\right)}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^1\left(xe^x+\sin\frac{\pi x}{4}\right)dx=1+\frac4\pi\left(1-\frac{\sqrt2}{2}\right)}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: Tests the application of integration by parts, standard trigonometric integration and evaluation of definite integrals.
JEE Main: Develops the ability to identify and combine different integration techniques in one problem.
JEE Advanced: Reinforces the use of integration by parts and exact trigonometric values in definite integral evaluation.
Conceptual Understanding: Demonstrates the linearity property of integration, which allows a sum of functions to be integrated term by term.

🚫 Common Mistakes

Common Mistakes

›

Incorrect integration by parts: Remember that \(\int xe^x\,dx=e^x(x-1)+C\).
Incorrect choice of \(u\) and \(dv\): Choosing \(u=x\) and \(dv=e^x\,dx\) makes the calculation straightforward.
Missing the reciprocal coefficient: The integral of \(\sin(ax)\) contains the factor \(\frac1a\).
Sign error: The antiderivative of sine is negative cosine, not positive cosine.
Incorrect trigonometric value: Use \(\cos\frac{\pi}{4}=\frac{\sqrt2}{2}\) and \(\cos0=1\).
Incorrect limit evaluation: Always subtract the value at the lower limit from the value at the upper limit.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Integration by parts is useful for integrating products of algebraic and exponential functions.
The integral of \(xe^x\) is \(e^x(x-1)+C\).
The integral of \(\sin(ax)\) is \(-\frac{\cos(ax)}{a}+C\).
Integration is linear, so the integral of a sum can be evaluated as the sum of individual integrals.
Careful application of definite limits is essential for obtaining the correct result.

📘 Concept & Theory

Theory and Concept

›

This question is based on the standard integration formula for the inverse tangent function.

\[ \int\frac{dx}{1+x^2}=\tan^{-1}x+C \]

For a definite integral, the Fundamental Theorem of Calculus gives:

\[ \int\limits_a^b f(x)\,dx=F(b)-F(a) \]

where \(F'(x)=f(x)\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the standard inverse tangent integral.
Write its antiderivative.
Substitute the upper and lower limits.
Use the standard inverse tangent values.
Match the result with the given options.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 5 steps

Given —
\[I=\int\limits_1^{\sqrt3}\frac{dx}{1+x^2}\]
Apply the standard integration formula
We know that:
\[\int\frac{dx}{1+x^2}=\tan^{-1}x+C\]
Therefore,
\[I=\left[\tan^{-1}x\right]_1^{\sqrt3}\]
Apply the limits of integration
\[\begin{aligned}I&=\tan^{-1}\sqrt3-\tan^{-1}1\end{aligned}\]
Substitute the standard inverse tangent values
We know that:
\[\tan^{-1}\sqrt3=\frac{\pi}{3}\]

\[\tan^{-1}1=\frac{\pi}{4}\]
Substituting:
\[ \begin{aligned} I &=\frac{\pi}{3}-\frac{\pi}{4}\\ &=\frac{4\pi-3\pi}{12}\\ &=\frac{\pi}{12} \end{aligned} \]

💡 Answer

Final Answer

›

\[ \boxed{I=\frac{\pi}{12}} \]

Correct option: (D) \(\dfrac{\pi}{12}\)

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: Tests the direct application of a standard integration formula and definite limits.
JEE Main: Tests familiarity with standard inverse trigonometric values and quick evaluation.
JEE Advanced: Reinforces the relationship between definite integration and inverse trigonometric functions.

🚫 Common Mistakes

Common Mistakes

›

Do not interchange the upper and lower limits.
Remember that \(\tan^{-1}1=\frac{\pi}{4}\), not \(\frac{\pi}{3}\).
Use a common denominator when subtracting the two fractions.
Do not confuse \(\tan^{-1}\sqrt3\) with \(\tan^{-1}3\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

The standard integral of \(\frac{1}{1+x^2}\) is \(\tan^{-1}x+C\).
\(\tan^{-1}\sqrt3=\frac{\pi}{3}\).
\(\tan^{-1}1=\frac{\pi}{4}\).
A definite integral is evaluated by subtracting the antiderivative at the lower limit from its value at the upper limit.

📘 Concept & Theory

Theory and Concept

›

This problem is based on the standard inverse tangent integration formula:

\[ \int\frac{dx}{a^2+b^2x^2} = \frac{1}{ab}\tan^{-1}\left(\frac{bx}{a}\right)+C \]

The formula is obtained by expressing the denominator in the form required for inverse tangent integration.

In the given integral:

\[ a=2,\qquad b=3 \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the constants \(a\) and \(b\).
Apply the standard inverse tangent integration formula.
Substitute the upper and lower limits.
Evaluate the inverse tangent values.
Compare the result with the given options.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 5 steps

Given —
\[I=\int\limits_0^{\frac23}\frac{dx}{4+9x^2}\]
Identify the standard form
Comparing the denominator with \(a^2+b^2x^2\), we obtain:
\[a^2=4\implies a=2\]

\[b^2=9\implies b=3\]
Apply the standard integration formula
\[\begin{aligned}I&=\left[\frac{1}{2\times3}\tan^{-1}\left(\frac{3x}{2}\right)\right]_0^{\frac23}\\ &=\frac16\left[\tan^{-1}\left(\frac{3x}{2}\right)\right]_0^{\frac23}\end{aligned}\]
Apply the limits of integration
\[\begin{aligned}I&=\frac16\left[\tan^{-1}\left(\frac{3}{2}\times\frac23\right)-\tan^{-1}\left(\frac{3}{2}\times0\right)\right]\\ &=\frac16\left[\tan^{-1}(1)-\tan^{-1}(0)\right]\end{aligned}\]
Substitute the standard inverse tangent values
We know that:
\[\tan^{-1}(1)=\frac{\pi}{4}\]

\[\tan^{-1}(0)=0\]
Therefore,
\[\begin{aligned}I&=\frac16\left(\frac{\pi}{4}-0\right)\\ &=\frac{\pi}{24}\end{aligned}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\limits_0^{\frac23}\frac{dx}{4+9x^2}=\frac{\pi}{24}}\]

Correct option: (C) \(\dfrac{\pi}{24}\)

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: Tests the direct application of the standard inverse tangent integration formula.
JEE Main: Develops speed and accuracy in identifying standard integral forms.
JEE Advanced: Reinforces the relationship between algebraic expressions and inverse trigonometric functions.

🚫 Common Mistakes

Common Mistakes

›

Do not confuse \(a^2=4\) with \(a=4\). The correct value is \(a=2\).
Do not omit the coefficient \(\frac{1}{ab}=\frac16\).
Substitute the upper limit carefully: \(\frac32\times\frac23=1\).
Remember that \(\tan^{-1}(1)=\frac{\pi}{4}\).
Always subtract the lower-limit value from the upper-limit value.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

The integral of \(\frac{1}{a^2+b^2x^2}\) produces an inverse tangent expression.
The coefficient outside the inverse tangent function is \(\frac{1}{ab}\).
Correct identification of constants simplifies definite integration.
Standard inverse tangent values are frequently used in objective examination questions.

Integrals

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Integrals

Chapter Complete!

Recent posts

INTEGRALS — Learning Resources

Frequently Asked Questions

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