NCERT Class 12 Maths Integrals Misc Exercise Solution

📘 Concept & Theory

Theory and Concept

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This problem involves the method of integration by partial fractions, which is particularly useful when the integrand is a rational function (a ratio of two polynomials).

The given integrand has a cubic polynomial in the denominator. To integrate it, we first factorise the denominator into linear factors and then express the rational function as a sum of simpler fractions.

Important concepts and formulas:

1. Factorisation of the denominator:

\[ x-x^3=x(1-x^2)=x(1-x)(1+x) \]

2. Partial fraction decomposition for three distinct linear factors:

\[ \frac{1}{x(x+1)(1-x)} = \frac{A}{x}+\frac{B}{x+1}+\frac{C}{1-x} \]

3. Standard integration formulas:

\[ \int\frac{dx}{x}=\ln|x|+C \]

\[ \int\frac{dx}{a+bx} = \frac{1}{b}\ln|a+bx|+C \]

Important observation: While integrating \(\frac{1}{1-x}\), the derivative of \(1-x\) is \(-1\). Therefore, a negative sign appears in its integral.

🗺️ Solution Roadmap

Step-by-step Plan

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Factorise the denominator \(x-x^3\) into three distinct linear factors.
Express the resulting rational function in partial fractions by introducing constants \(A\), \(B\), and \(C\).
Multiply both sides by the common denominator and compare coefficients of corresponding powers of \(x\).
Determine the values of the unknown constants.
Substitute the constants into the partial fraction expression.
Integrate each simpler fraction separately using standard logarithmic integration formulas.
Combine the logarithmic terms and include the arbitrary constant of integration.

✏️ Solution

Complete Solution

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Step-by-step Solution · 27 steps

Given —
\[I=\int\frac{dx}{x-x^3}\]
Factorise the denominator
Taking \(x\) common from the denominator, we get:
\[I=\int\frac{dx}{x(1-x^2)}\]
Using the identity \(a^2-b^2=(a-b)(a+b)\), we have:
\[1-x^2=(1-x)(1+x)\]
Therefore,
\[I=\int\frac{dx}{x(1+x)(1-x)}\]
Resolve into partial fractions
Since the denominator consists of three distinct linear factors, we assume:
\[\frac{1}{x(1+x)(1-x)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{1-x}\]
Taking the LCM on the right-hand side, we obtain:
\[\frac{1}{x(1+x)(1-x)}=\frac{A(1+x)(1-x)+Bx(1-x)+Cx(x+1)}{x(1+x)(1-x)}\]
Equating the numerators,
\[1=A(1-x^2)+Bx(1-x)+Cx(x+1)\]
Expanding the terms,
\[1=A-Ax^2+Bx-Bx^2+Cx+Cx^2\]
Rearranging and collecting the coefficients of like powers of \(x\),
\[1=A+(B+C)x+(-A-B+C)x^2\]
Compare the coefficients.
Comparing the coefficients of corresponding powers of \(x\) on both sides, we get the following equations:
Comparing the constant terms

\[A=1 \tag{1}\]
Comparing the coefficients of \(x\):

\[B+C=0 \tag{2}\]
Comparing the coefficients of \(x^2\):

\[-A-B+C=0\tag{3}\]
Calculate the values of \(A\), \(B\), and \(C\)
From equation (1),
\[A=1\]
Substituting \(A=1\) in equation (3),
\[-1-B+C=0\]

\[C-B=1\tag{4}\]
We already have equation (2):
\[B+C=0\tag{4}\]
Adding equations (2) and (4),
\[ \begin{aligned} &(C-B)+(B+C)=1+0\\ &C-B+B+C=1\\ &2C=1\\ &\boxed{C=\frac{1}{2}} \end{aligned} \]
Substituting \(C=\frac{1}{2}\) in equation (2),
\[ \begin{aligned} B+C&=0\\ B+\frac{1}{2}&=0\\ B&=-\frac{1}{2} \end{aligned} \]
Thus, the required constants are:
\[\boxed{A=1,\quad B=-\frac{1}{2},\quad C=\frac{1}{2}}\]
Substitute the constants into the partial fractions
Substituting the values of \(A\), \(B\), and \(C\), we obtain:
\[\frac{1}{x(1+x)(1-x)}=\frac{1}{x}-\frac{1}{2(x+1)}+\frac{1}{2(1-x)}\]
Therefore, the original integral becomes:
\[I=\int\left[\frac{1}{x}-\frac{1}{2(x+1)}+\frac{1}{2(1-x)}\right]dx\]
Integrate each term separately
Using the linearity property of integration,
\[I=\int\frac{dx}{x}-\frac{1}{2}\int\frac{dx}{x+1}+\frac{1}{2}\int\frac{dx}{1-x}\]
Now, using the standard integration formulas:
\[ \int\frac{dx}{x}=\ln|x| \]

\[ \int\frac{dx}{x+1}=\ln|x+1| \]
For the third integral, put \(u=1-x\). Then,
\[u=1-x,\quad du=-dx\]
Therefore,
\[ \begin{aligned} \int\frac{dx}{1-x} &=-\int\frac{du}{u}\\ &=-\ln|u|\\ &=-\ln|1-x| \end{aligned} \]
Substituting these results into the original expression,
\[ \begin{aligned} I &=\ln|x| -\frac{1}{2}\ln|x+1| -\frac{1}{2}\ln|1-x|+C\\ &=\ln|x| -\frac{1}{2} \left[\ln|x+1|+\ln|1-x|\right]+C \end{aligned} \]
Using the logarithmic identity:
\[\ln a+\ln b=\ln(ab)\]
we get:
\[ \begin{aligned} I &=\ln|x|-\frac{1}{2}\ln|(x+1)(1-x)|+C\\ &=\ln|x|-\frac{1}{2}\ln|1-x^2|+C \end{aligned} \]
Hence, the required integral is:
\[\boxed{\displaystyle\int\frac{dx}{x-x^3}=\ln|x|-\frac{1}{2}\ln|1-x^2|+C}\]
Equivalently, the answer can be expressed as:
\[\boxed{\displaystyle\ln|x|+\frac{1}{2}\ln\left|\frac{1}{1-x^2}\right|+C}\]

💡 Answer

Final Answer

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\[\boxed{\displaystyle\ln|x|+\frac{1}{2}\ln\left|\frac{1}{1-x^2}\right|+C}\]

🎯 Exam Significance

Exam Significance

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This problem demonstrates the application of partial fraction decomposition in evaluating rational functions.
It tests factorisation of polynomial denominators and the systematic determination of unknown constants.
It reinforces the use of standard logarithmic integration formulas, particularly the importance of signs when integrating linear expressions.
Students should show the coefficient-comparison steps clearly to secure marks for the method, even if an arithmetic error occurs later.

For Competitive Entrance Examinations (JEE Main, JEE Advanced, CUET and other engineering entrance tests):

Partial fraction decomposition is an important technique for evaluating rational integrals.
The ability to identify suitable linear factors and determine coefficients efficiently is useful in both single-correct and multiple-correct questions.
Recognising logarithmic derivatives helps reduce lengthy calculations and improves accuracy.
Verification by differentiation provides a reliable way to identify sign errors in indefinite integration.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 6 points

Factorisation first: Always factorise the denominator completely before applying partial fractions.
Correct decomposition: For distinct linear factors, assign a separate constant numerator to each factor.
Coefficient comparison: Compare coefficients of corresponding powers of \(x\) to determine the unknown constants systematically.
Watch the negative sign: Remember that \(\int\frac{dx}{1-x}=-\ln|1-x|+C\).
Use logarithmic identities: Combining logarithmic terms can produce a shorter and more elegant final expression.
Always include the integration constant: An indefinite integral must contain the arbitrary constant \(C\).

📘 Concept & Theory

Theory and Concept

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This problem involves the technique of rationalisation using conjugate surds, followed by substitution and standard integration.

When the denominator of an integrand contains the sum of two square roots, direct integration is generally difficult. We can simplify the expression by multiplying the numerator and denominator by the conjugate of the denominator.

Important algebraic identity:

\[ (A+B)(A-B)=A^2-B^2 \]

In the given problem, the conjugate of \(\sqrt{x+a}+\sqrt{x+b}\) is \(\sqrt{x+a}-\sqrt{x+b}\).

We also use the standard integration formula:

\[ \int\frac{du}{\sqrt{u}}=2\sqrt{u}+C \]

🗺️ Solution Roadmap

Step-by-step Plan

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Write the given integral and identify the conjugate of the denominator.
Multiply the numerator and denominator by the conjugate surd.
Apply the difference of squares identity to simplify the denominator.
Recognise that the resulting numerator is the derivative of a suitable expression involving square roots.
Integrate the simplified expression term by term.
Substitute the original variables and include the arbitrary constant of integration.

✏️ Solution

Complete Solution

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Step-by-step Solution · 10 steps

Given —
\[I=\int\frac{dx}{\sqrt{x+a}+\sqrt{x+b}}\]
Rationalise the denominator.
Multiplying and dividing the integrand by the conjugate expression \(\sqrt{x+a}-\sqrt{x+b}\), we obtain:
\[I=\int\frac{\sqrt{x+a}-\sqrt{x+b}}{[\sqrt{x+a}+\sqrt{x+b}][\sqrt{x+a}-\sqrt{x+b}]}\,dx\]
Using the identity \((A+B)(A-B)=A^2-B^2\), the denominator becomes:
\[ \begin{aligned} &(\sqrt{x+a})^2-(\sqrt{x+b})^2\\ &=(x+a)-(x+b)\\ &=a-b \end{aligned} \]
Therefore,
\[I=\int\frac{\sqrt{x+a}-\sqrt{x+b}}{a-b}\,dx\]
Taking the constant denominator outside the integral,
\[I=\frac{1}{a-b}\left[\int\sqrt{x+a}\,dx-\int\sqrt{x+b}\,dx\right]\]
Integrate the first term
Using the substitution:
\[u=x+a\]

\[du=dx\]
We get:
\[ \begin{aligned} \int\sqrt{x+a}\,dx &=\int u^{1/2}\,du\\ &=\frac{u^{3/2}}{3/2}+C\\ &=\frac{2}{3}u^{3/2}\\ &=\frac{2}{3}(x+a)^{3/2} \end{aligned} \]
Integrate the second term
Similarly, putting:
\[v=x+b\]

\[dv=dx\]
We obtain:
\[ \begin{aligned} \int\sqrt{x+b}\,dx &=\int v^{1/2}\,dv\\ &=\frac{v^{3/2}}{3/2}\\ &=\frac{2}{3}v^{3/2}\\ &=\frac{2}{3}(x+b)^{3/2} \end{aligned} \]
Substitute both results into the original integral
\[\begin{aligned}I&=\frac{1}{a-b}\left[\frac{2}{3}(x+a)^{3/2}-\frac{2}{3}(x+b)^{3/2}\right]+C\\ &=\frac{2}{3(a-b)}\left[(x+a)^{3/2}-(x+b)^{3/2}\right]+C\end{aligned}\]
Hence, the required integral is:
\[\boxed{\displaystyle\int\frac{dx}{\sqrt{x+a}+\sqrt{x+b}}=\frac{2}{3(a-b)}\left[(x+a)^{3/2}-(x+b)^{3/2}\right]+C}\]
Here, \(a\ne b\), and the expression is considered on an interval where the square roots are defined.

💡 Answer

Final Answer

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\[\boxed{\displaystyle\int\frac{dx}{\sqrt{x+a}+\sqrt{x+b}}=\frac{2}{3(a-b)}\left[(x+a)^{3/2}-(x+b)^{3/2}\right]+C}\]

🎯 Exam Significance

Exam Significance

›

This problem tests the application of algebraic identities in simplifying integrands containing surds.
It demonstrates how rationalisation can transform a complicated-looking integral into elementary power integrals.
Students should clearly show the conjugate multiplication and the difference of squares simplification to obtain method marks.
It reinforces the power rule of integration and the importance of including the constant of integration.

For Competitive Entrance Examinations (JEE Main, JEE Advanced, CUET and other entrance tests):

Recognising conjugate expressions is an important algebraic skill in integration problems.
This technique is useful for simplifying integrals involving sums or differences of surds.
The problem develops the ability to identify hidden elementary integrals after algebraic manipulation.
Understanding the special case \(a=b\) strengthens awareness of parameter restrictions and prevents division-by-zero errors.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 6 points

Identify the conjugate: For a denominator containing the sum of two square roots, multiply by their difference.
Apply the difference of squares identity: The expression \((x+a)-(x+b)\) simplifies to the constant \(a-b\).
Convert surds into powers: Write \(\sqrt{x+a}\) as \((x+a)^{1/2}\) to apply the power rule of integration.
Use linearity: Integrate the resulting terms separately.
Check parameter restrictions: The main result requires \(a\ne b\). Handle \(a=b\) separately.
Verify by differentiation: Differentiating the final expression must reproduce the original integrand.

📘 Concept & Theory

Theory and Concept

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This problem involves integration by substitution, particularly the use of the substitution \(x=\frac{a}{t}\) to simplify an integrand containing a quadratic expression under a square root.

The expression under the square root can be factorised as:

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

However, directly integrating the given expression is not straightforward. A suitable substitution transforms the quadratic expression into a difference of squares, which can then be integrated using partial fractions.

Important concepts and formulas:

1. Substitution:

\[ x=\frac{a}{t} \]

\[ dx=-\frac{a}{t^2}\,dt \]

2. Difference of squares:

\[ t^2-1=(t-1)(t+1) \]

3. Standard logarithmic integration formula:

\[ \int\frac{dt}{t^2-1} = \frac{1}{2}\ln\left| \frac{t-1}{t+1} \right|+C \]

Important observation: The factor \(\frac{1}{2}\) in the logarithmic integration formula is essential. It arises from resolving \(\frac{1}{t^2-1}\) into partial fractions.

🗺️ Solution Roadmap

Step-by-step Plan

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Consider the given integral and identify a suitable substitution.
Put \(x=\frac{a}{t}\) and calculate \(dx\).
Express the square-root term \(\sqrt{ax-x^2}\) in terms of \(t\).
Substitute the transformed expressions into the original integral and simplify.
Resolve the resulting rational function into partial fractions.
Integrate using the standard logarithmic formula.
Replace \(t\) by \(\frac{a}{x}\) and simplify the final expression.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Given —
\[I=\int\frac{dx}{x\sqrt{ax-x^2}}\]
Let us use the substitution:
\[x=\frac{a}{t}\]
Differentiating both sides with respect to \(t\),
\[\frac{dx}{dt}=-\frac{a}{t^2}\]
Therefore,
\[dx=-\frac{a}{t^2}\,dt\]
Express the square-root term in terms of \(t\).
We have:
\[ \begin{aligned} \sqrt{ax-x^2} &=\sqrt{a\left(\frac{a}{t}\right) -\left(\frac{a}{t}\right)^2}\\ &=\sqrt{\frac{a^2}{t}-\frac{a^2}{t^2}}\\ &=\sqrt{\frac{a^2t-a^2}{t^2}}\\ &=\sqrt{\frac{a^2(t-1)}{t^2}}\\ &=\frac{|a|}{|t|}\sqrt{t-1} \end{aligned} \]
For the usual textbook assumption \(a>0\) and \(t>0\), this becomes:
\[\sqrt{ax-x^2}=\frac{a}{t}\sqrt{t-1}\]
Substitute the expressions into the integral
Recall that:
\[x=\frac{a}{t},\quad dx=-\frac{a}{t^2}\,dt\]
Substituting these expressions,
\[ \begin{aligned} I &=\int \frac{-\frac{a}{t^2}\,dt} {\frac{a}{t}\cdot\frac{a}{t}\sqrt{t-1}}\\ &=-\frac{1}{a} \int\frac{dt}{\sqrt{t-1}} \end{aligned} \]
Integrate using the power rule
Put:
\[u=t-1\]

\[du=dt\]
Therefore,
\[ \begin{aligned} I &=-\frac{1}{a}\int u^{-1/2}\,du\\ &=-\frac{1}{a} \left(\frac{u^{1/2}}{1/2}\right)+C\\ &=-\frac{2}{a}\sqrt{u}+C\\ &=-\frac{2}{a}\sqrt{t-1}+C \end{aligned} \]
Substitute the original variable
Since:
\[t=\frac{a}{x}\]
we obtain:
\[ \begin{aligned} I &=-\frac{2}{a} \sqrt{\frac{a}{x}-1}+C\\ &=-\frac{2}{a} \sqrt{\frac{a-x}{x}}+C \end{aligned} \]
Hence, the required integral is:
\[\boxed{\displaystyle\int\frac{dx}{x\sqrt{ax-x^2}}=-\frac{2}{a}\sqrt{\frac{a-x}{x}}+C}\]

💡 Answer

Final Answer

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\[\boxed{\displaystyle\int\frac{dx}{x\sqrt{ax-x^2}}=-\frac{2}{a}\sqrt{\frac{a-x}{x}}+C}\]

🎯 Exam Significance

Exam Significance

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This problem illustrates how a carefully selected substitution can simplify an integral involving a quadratic expression under a square root.
It tests differentiation of substituted expressions and the correct transformation of the differential \(dx\).
It reinforces the power rule of integration and the importance of retaining the negative sign generated by the substitution.
Writing each substitution and simplification step explicitly helps students demonstrate mathematical reasoning and secure method marks.

For Competitive Entrance Examinations (JEE Main, JEE Advanced, CUET and other entrance tests):

Choosing an appropriate substitution is an important problem-solving skill in integration.
Expressions involving quadratic surds frequently require algebraic transformations before standard integration techniques can be applied.
The problem develops familiarity with substitutions that convert complicated radical expressions into simpler power functions.
Verification through differentiation is useful for detecting errors involving negative signs, powers and numerical coefficients.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 5 points

Choose a suitable substitution: The substitution \(x=\frac{a}{t}\) transforms the quadratic expression into a simpler form.
Transform the differential carefully: Always calculate \(dx\) before substituting into the integral.
Simplify the radical correctly: Under the assumption \(a>0\) and \(t>0\), \(\sqrt{ax-x^2}=\frac{a}{t}\sqrt{t-1}\).
Apply the power rule: The transformed integral can be evaluated directly using the standard power formula.
Watch the negative sign: The negative sign in \(dx=-\frac{a}{t^2}dt\) must be retained throughout the calculation.

📘 Concept & Theory

Theory and Concept

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This problem involves integration by substitution and the application of the power rule of integration. The key is to recognise that the expression \(x^4+1\) can be transformed into a simpler expression by choosing a suitable substitution.

We observe that the denominator contains \(x^2\) and a fractional power of \(x^4+1\). Dividing the expression inside the bracket by \(x^4\) suggests the substitution:

\[ t=\frac{1}{x} \]

This substitution transforms the expression into a power function involving \(1+t^4\), which can be integrated directly.

Important concepts and formulas:

1. Reciprocal substitution:

\[ t=\frac{1}{x} \]

\[ x=\frac{1}{t}, \qquad dx=-\frac{1}{t^2}\,dt \]

2. Laws of indices:

\[ (a^m)^n=a^{mn} \]

\[ \frac{a^m}{a^n}=a^{m-n} \]

3. Power rule of integration:

\[ \int u^n\,du=\frac{u^{n+1}}{n+1}+C, \qquad n\ne-1 \]

🗺️ Solution Roadmap

Step-by-step Plan

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Consider the given integral and identify a suitable reciprocal substitution.
Put \(t=\frac{1}{x}\) and calculate \(dx\).
Express \(x^2\) and \(x^4+1\) in terms of \(t\).
Simplify the resulting fractional powers carefully using the laws of indices.
Reduce the transformed integral to a standard power integral.
Integrate and substitute the original variable.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 20 steps

Given —
\[I=\int\frac{dx}{x^2(x^4+1)^{3/4}}\]
Let us put:
\[t=\frac{1}{x}\]
Therefore,
\[x=\frac{1}{t}\]
Differentiating both sides with respect to \(t\),
\[\frac{dx}{dt}=-\frac{1}{t^2}\]
Hence,
\[dx=-\frac{1}{t^2}\,dt\]
Express the denominator in terms of \(t\)
Since \(x=\frac{1}{t}\), we have:
\[x^2=\frac{1}{t^2}\]
Also,
\[ \begin{aligned} x^4+1 &=\frac{1}{t^4}+1\\ &=\frac{1+t^4}{t^4} \end{aligned} \]
Therefore,
\[ \begin{aligned} (x^4+1)^{3/4} &=\left(\frac{1+t^4}{t^4}\right)^{3/4}\\ &=\frac{(1+t^4)^{3/4}}{t^3} \end{aligned} \]
Here, we have used \(t>0\) for the displayed simplification. More generally, the denominator is \(|t|^3\).
Substitute into the original integral
\[ I=\int \frac{-\frac{1}{t^2}\,dt} {\frac{1}{t^2}\cdot \frac{(1+t^4)^{3/4}}{t^3}} \]
Simplifying the denominator,
\[ \begin{aligned} I &=-\int \frac{\frac{1}{t^2}\,dt} {\frac{(1+t^4)^{3/4}}{t^5}}\\ &=-\int \frac{t^3}{(1+t^4)^{3/4}}\,dt \end{aligned} \]
Apply another substitution
Put:
\[u=1+t^4\]
Differentiating both sides,
\[du=4t^3\,dt\]
Therefore,
\[t^3\,dt=\frac{du}{4}\]
Substituting these expressions,
\[ \begin{aligned} I &=-\int\frac{t^3}{(1+t^4)^{3/4}}\,dt\\ &=-\frac{1}{4}\int u^{-3/4}\,du \end{aligned} \]
Integrate using the power rule
Applying the standard formula:
\[\int u^n\,du=\frac{u^{n+1}}{n+1}+C\]
we get:
\[ \begin{aligned} I &=-\frac{1}{4} \left(\frac{u^{-3/4+1}}{-3/4+1}\right)+C\\ &=-\frac{1}{4} \left(\frac{u^{1/4}}{1/4}\right)+C\\ &=-u^{1/4}+C \end{aligned} \]
Substituting \(u=1+t^4\),
\[I=-(1+t^4)^{1/4}+C\]
Substitute the original variable
Since \(t=\frac{1}{x}\),
\[ \begin{aligned} I &=-\left(1+\frac{1}{x^4}\right)^{1/4}+C\\ &=-\left(\frac{x^4+1}{x^4}\right)^{1/4}+C \end{aligned} \]
Using \((x^4)^{1/4}=|x|\), we obtain:
\[I=-\frac{(x^4+1)^{1/4}}{|x|}+C\]
Hence, the required integral is:
\[\boxed{\displaystyle\int\frac{dx}{x^2(x^4+1)^{3/4}}=-\frac{\sqrt[4]{x^4+1}}{|x|}+C}\]
On the usual textbook interval \(x>0\), this can be written more simply as:
\[\boxed{\displaystyle I=-\frac{\sqrt[4]{x^4+1}}{x}+C}\]

💡 Answer

Final Answer

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\[\boxed{\displaystyle I=-\frac{\sqrt[4]{x^4+1}}{x}+C}\]

🎯 Exam Significance

Exam Significance

›

This problem demonstrates the importance of selecting an appropriate substitution for integrals involving fractional powers and algebraic expressions.
It tests the correct transformation of variables, differentials and fractional powers.
It reinforces the application of the power rule after simplifying a complicated integrand.
Students should show the intermediate algebraic simplifications clearly, particularly when converting \(x^4+1\) into an expression involving the substituted variable.

For Competitive Entrance Examinations (JEE Main, JEE Advanced, CUET and other entrance tests):

Recognising reciprocal substitutions is an important integration skill.
The problem develops the ability to simplify expressions containing negative powers and fractional indices.
It illustrates how a complicated-looking rational expression can be converted into a straightforward power integral.
Understanding the domain and absolute-value considerations helps avoid errors when working with real-valued fractional powers.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 6 points

Recognise the reciprocal structure: The combination of \(x^2\) and \(x^4+1\) suggests trying \(t=\frac{1}{x}\).
Transform the differential carefully: Remember that \(dx=-\frac{1}{t^2}\,dt\).
Apply the laws of indices: Simplify fractional powers systematically rather than skipping algebraic steps.
Use a second substitution: The expression \(1+t^4\) suggests \(u=1+t^4\), whose differential contains the required factor \(t^3\,dt\).
Apply the power rule: The transformed integral becomes an elementary power integral.
Consider absolute values: Since \((x^4)^{1/4}=|x|\), the general real-valued antiderivative contains \(|x|\) in the denominator.

📘 Concept & Theory

Theory and Concept

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This problem involves integration by substitution followed by partial fraction decomposition. The integrand contains fractional powers of \(x\), namely \(x^{1/2}\) and \(x^{1/3}\). Such expressions can often be simplified by choosing a substitution that eliminates the fractional exponents.

The denominators contain powers with denominators 2 and 3. Their least common multiple is 6. Therefore, we choose:

\[ t=x^{1/6} \]

Consequently,

\[ x=t^6,\qquad dx=6t^5\,dt \]

The fractional powers become ordinary integer powers:

\[ x^{1/2}=t^3,\qquad x^{1/3}=t^2 \]

This transforms the original integral into a rational function of \(t\), which can be evaluated using partial fractions.

Important concepts and formulas:

1. Least common multiple of fractional exponents:

\[ \operatorname{LCM}(2,3)=6 \]

2. Substitution:

\[ t=x^{1/6},\qquad x=t^6 \]

\[ dx=6t^5\,dt \]

3. Factorisation:

\[ t^3+t^2=t^2(t+1) \]

4. Standard logarithmic integration formula:

\[ \int\frac{dt}{t+1}=\ln|t+1|+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the fractional powers in the given integrand.
Choose \(t=x^{1/6}\), using the LCM of the denominators of the fractional exponents.
Express \(x\), \(dx\), \(x^{1/2}\) and \(x^{1/3}\) in terms of \(t\).
Substitute these expressions into the original integral.
Factorise the resulting polynomial and simplify the rational function.
Integrate the simplified expression using standard formulas.
Substitute \(t=x^{1/6}\) to obtain the final answer.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 17 steps

Given —
\[ I=\int\frac{dx}{x^{1/2}+x^{1/3}}\]
Since the fractional powers have denominators 2 and 3, their LCM is 6. Therefore, put:
\[t=x^{1/6}\]
Raising both sides to the sixth power,
\[x=t^6\]
Differentiating with respect to \(t\),
\[\frac{dx}{dt}=6t^5\]
Hence,
\[dx=6t^5\,dt\]
Express the fractional powers in terms of \(t\).
We have:
\[ \begin{aligned} x^{1/2} &=(t^6)^{1/2}\\ &=t^3 \end{aligned} \]
Similarly,
\[ \begin{aligned} x^{1/3} &=(t^6)^{1/3}\\ &=t^2 \end{aligned} \]
Therefore, the denominator becomes:
\[x^{1/2}+x^{1/3}=t^3+t^2\]
Substitute into the original integral
Substituting \(x=t^6\) and \(dx=6t^5\,dt\), we get:
\[ \begin{aligned} I &=\int\frac{6t^5}{t^3+t^2}\,dt\\ &=6\int\frac{t^5}{t^2(t+1)}\,dt\\ &=6\int\frac{t^3}{t+1}\,dt \end{aligned} \]
Simplify the rational function using polynomial division
We divide \(t^3\) by \(t+1\).
\[\begin{aligned}t^3&=(t+1)(t^2-t+1)-1\end{aligned}\]
Therefore,
\[\frac{t^3}{t+1}=t^2-t+1-\frac{1}{t+1}\]
Substituting this expression into the integral,
\[I=6\int\left(t^2-t+1-\frac{1}{t+1}\right)dt\]
Integrate each term separately
Using the linearity property of integration,
\[ \begin{aligned} I=6\bigg[ &\int t^2\,dt-\int t\,dt\\ &+\int 1\,dt-\int\frac{1}{t+1}\,dt \bigg] \end{aligned} \]
Now, applying the standard integration formulas:
\[\int t^2\,dt=\frac{t^3}{3}\]

\[\int t\,dt=\frac{t^2}{2}\]

\[\int 1\,dt=t\]

\[\int\frac{1}{t+1}\,dt=\ln|t+1|\]
Thus,
\[\begin{aligned}I&=6\left[\frac{t^3}{3}-\frac{t^2}{2}+t-\ln|t+1|\right]+C\\ &=2t^3-3t^2+6t-6\ln|t+1|+C\end{aligned}\]
Substitute the original variable
Since \(t=x^{1/6}\), we have:
\[t^3=x^{1/2},\quad t^2=x^{1/3}\]
Substituting these expressions,
\[ \begin{aligned} I={}&2x^{1/2}-3x^{1/3} +6x^{1/6}\\ &-6\ln|x^{1/6}+1|+C \end{aligned} \]
Hence, the required integral is:
\[ \boxed{ \begin{aligned} \displaystyle \int\frac{dx}{x^{1/2}+x^{1/3}} ={}&2\sqrt{x}-3\sqrt[3]{x}\\ &+6\sqrt[6]{x} -6\ln(1+\sqrt[6]{x})+C \end{aligned} } \]
For the usual real-valued textbook domain \(x>0\), the absolute-value signs around \(1+\sqrt[6]{x}\) are unnecessary because this expression is positive.

💡 Answer

Final Answer

›

\[ \boxed{ \begin{aligned} \displaystyle \int\frac{dx}{x^{1/2}+x^{1/3}} ={}&2\sqrt{x}-3\sqrt[3]{x}\\ &+6\sqrt[6]{x} -6\ln(1+\sqrt[6]{x})+C \end{aligned} } \]

🎯 Exam Significance

Exam Significance

›

This problem demonstrates how to eliminate fractional powers by choosing a substitution based on the LCM of their denominators.
It tests the conversion of a fractional-power integral into a rational algebraic expression.
It reinforces polynomial division and the integration of polynomial and logarithmic terms.
Showing the substitution, factorisation and polynomial division in detail helps establish a clear mathematical method and secure stepwise marks.

For Competitive Entrance Examinations (JEE Main, JEE Advanced, CUET and other entrance tests):

Recognising the LCM-based substitution is a useful technique for integrals involving multiple fractional exponents.
The problem develops algebraic fluency by converting fractional powers into integer powers.
Polynomial division is an important simplification technique for rational functions arising after substitution.
The ability to recognise a logarithmic term in the resulting expression is useful in evaluating indefinite integrals efficiently.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Find the LCM: For fractional powers with denominators 2 and 3, choose a substitution involving their LCM, 6.
Eliminate fractional exponents: The substitution \(t=x^{1/6}\) converts \(x^{1/2}\) and \(x^{1/3}\) into \(t^3\) and \(t^2\), respectively.
Factorise before simplifying: Use \(t^3+t^2=t^2(t+1)\) to reduce the rational function.
Apply polynomial division: Express \(\frac{t^3}{t+1}\) as \(t^2-t+1-\frac{1}{t+1}\).
Recognise logarithmic integrals: The term \(\frac{1}{t+1}\) produces a natural logarithm.
Substitute back carefully: Replace every occurrence of \(t\) with \(x^{1/6}\).
Verify the result: Differentiation confirms the correctness of the antiderivative.

📘 Concept & Theory

Theory and Concept

›

This integral involves a rational algebraic function whose denominator is the product of a linear factor and an irreducible quadratic factor. We can simplify it using the method of partial fractions.

The general decomposition for a rational expression with denominator \((x+1)(x^2+9)\) is

\[ \frac{5x}{(x+1)(x^2+9)} =\frac{A}{x+1}+\frac{Bx+C}{x^2+9} \]

After finding the constants \(A\), \(B\), and \(C\), the resulting integrals can be evaluated using standard formulas:

\[ \int\frac{dx}{x}=\ln|x|+C \]

\[ \int\frac{x}{x^2+a^2}\,dx =\frac12\ln(x^2+a^2)+C \]

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

🗺️ Solution Roadmap

Step-by-step Plan

›

Express the rational function as a sum of partial fractions.
Determine the unknown constants by comparing coefficients.
Split the original integral into three standard integrals.
Evaluate each integral using substitution and the inverse tangent formula.
Combine the results to obtain the final answer.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 19 steps

Apply partial fraction decomposition
We have
\[\frac{5x}{(x+1)(x^2+9)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+9}\]
Taking the LCM and comparing numerators,
\[5x=A(x^2+9)+(Bx+C)(x+1)\]
Expanding the right-hand side,
\[5x=Ax^2+9A+Bx^2+Bx+Cx+C\]

\[5x=(A+B)x^2+(B+C)x+(9A+C)\]
Comparing coefficients of corresponding powers of \(x\), we obtain
\[A+B=0 \tag{1}\]

\[B+C=5 \tag{2}\]

\[9A+C=0 \tag{3}\]
Step 2: Find the values of \(A\), \(B\), and \(C\).
From equation (1),
\[B=-A \tag{4}\]
Substituting \(B=-A\) in equation (2),
\[-A+C=5\]

\[C=5+A \tag{5}\]
Substituting equation (5) in equation (3),
\[9A+5+A=0\]

\[10A=-5\]

\[\boxed{A=-\frac12}\]
Using \(B=-A\),
\[\boxed{B=\frac12}\]
Using \(C=5+A\),
\[C=5-\frac12\]

\[\boxed{C=\frac92}\]
Therefore, the correct partial fraction decomposition is
\[\boxed{\frac{5x}{(x+1)(x^2+9)}=-\frac{1}{2(x+1)}+\frac{x+9}{2(x^2+9)}}\]
Substitute the values and split the integral
\[I=\int\left[-\frac{1}{2(x+1)}+\frac{x}{2(x^2+9)}+\frac{9}{2(x^2+9)}\right]dx\]
Separating the terms,
\[I=-\frac12\int\frac{dx}{x+1}+\frac12\int\frac{x}{x^2+9}\,dx+\frac92\int\frac{dx}{x^2+9}\]
Evaluate the first integral
\[-\frac12\int\frac{dx}{x+1}=-\frac12\ln|x+1|\]
Evaluate the second integral using substitution.
Let
\[t=x^2+9\]
Differentiating both sides,
\[dt=2x\,dx\]

\[x\,dx=\frac12 dt\]
Therefore,
\[ \begin{aligned} \frac12\int\frac{x}{x^2+9}\,dx &=\frac14\int\frac{dt}{t}\\ &=\frac14\ln|t|\\ &=\frac14\ln(x^2+9) \end{aligned} \]
Evaluate the third integral using the inverse tangent formula
Using the standard result
\[ \int\frac{dx}{x^2+a^2} =\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C \]
where \(a=3\), we get
\[ \begin{aligned} \frac92\int\frac{dx}{x^2+9} &=\frac92\cdot\frac13 \tan^{-1}\left(\frac{x}{3}\right)\\ &=\frac32\tan^{-1}\left(\frac{x}{3}\right) \end{aligned} \]
Combine all the results
Adding the three integrated terms,
\[ \begin{aligned} I={}&-\frac12\ln|x+1| +\frac14\ln(x^2+9)\\ &+\frac32\tan^{-1}\left(\frac{x}{3}\right)+C \end{aligned} \]
Hence, the required integral is
\[ \boxed{ \begin{aligned} \int\frac{5x}{(x+1)(x^2+9)}\,dx ={}&-\frac12\ln|x+1|\\ &+\frac14\ln(x^2+9)\\ &+\frac32\tan^{-1}\left(\frac{x}{3}\right)+C \end{aligned} } \]

💡 Answer

Final Answer

›

\[ \boxed{ \begin{aligned} \int\frac{5x}{(x+1)(x^2+9)}\,dx ={}&-\frac12\ln|x+1|\\ &+\frac14\ln(x^2+9)\\ &+\frac32\tan^{-1}\left(\frac{x}{3}\right)+C \end{aligned} } \]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem demonstrates the application of partial fraction decomposition to integrate rational algebraic functions.
JEE Main: Questions frequently test the identification of suitable partial fractions and the correct use of logarithmic and inverse tangent integration formulas.
JEE Advanced: The problem reinforces algebraic manipulation, coefficient comparison, and the combination of different standard integration techniques.
CUET and other entrance examinations: Understanding this method helps solve both direct integration questions and problems involving equivalent algebraic expressions.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For a product of a linear factor and an irreducible quadratic factor, use the decomposition \(\frac{A}{x+1}+\frac{Bx+C}{x^2+9}\).
Always compare coefficients carefully; the constants in this problem are \(A=-\frac12\), \(B=\frac12\), and \(C=\frac92\).
The integral \(\int\frac{x}{x^2+a^2}\,dx\) is evaluated using the substitution \(t=x^2+a^2\).
The integral \(\int\frac{dx}{x^2+a^2}\) produces an inverse tangent function.
Differentiating the final expression is an effective way to verify an indefinite integral.

📘 Concept & Theory

Theory and Concept

›

This integral involves a ratio of two sine functions whose arguments differ by a constant. The key technique is to express \(\sin x\) in terms of \(\sin(x-a)\) and \(\cos(x-a)\) using the sine addition formula.

Recall the trigonometric identity:

\[ \sin(A+B)=\sin A\cos B+\cos A\sin B \]

Taking \(A=x-a\) and \(B=a\), we obtain

\[ \sin x=\sin(x-a)\cos a+\cos(x-a)\sin a \]

This converts the original integrand into a sum of a constant and a cotangent function, both of which can be integrated directly.

🗺️ Solution Roadmap

Step-by-step Plan

›

Expand \(\sin x\) using the sine addition formula.
Divide each term by \(\sin(x-a)\).
Separate the resulting expression into a constant and a cotangent term.
Integrate using the standard logarithmic formula for cotangent.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Apply the sine addition formula
We know that
\[\sin x=\sin[(x-a)+a]\]
Using the identity \(\sin(A+B)=\sin A\cos B+\cos A\sin B\),
\[\sin x=\sin(x-a)\cos a+\cos(x-a)\sin a\]
Substitute this expression into the integral
\[I=\int\frac{\sin(x-a)\cos a+\cos(x-a)\sin a}{\sin(x-a)}\,dx\]
Separating the terms,
\[I=\int\left[\cos a+\sin a\frac{\cos(x-a)}{\sin(x-a)}\right]dx\]
Using \(\cot\theta=\frac{\cos\theta}{\sin\theta}\),
\[I=\int[\cos a+\sin a\cot(x-a)]\,dx\]
Integrate term by term
Since \(a\) is a constant, \(\cos a\) and \(\sin a\) are constants with respect to \(x\).
\[I=\cos a\int dx+\sin a\int\cot(x-a)\,dx\]
We know that
\[\int\cot u\,du=\ln|\sin u|+C\]
Putting \(u=x-a\), we have \(du=dx\). Therefore,
\[\int\cot(x-a)\,dx=\ln|\sin(x-a)|\]
Hence,
\[I=x\cos a+\sin a\ln|\sin(x-a)|+C\]
\[\boxed{\int\frac{\sin x}{\sin(x-a)}\,dx=x\cos a+\sin a\ln|\sin(x-a)|+C}\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{\sin x}{\sin(x-a)}\,dx=x\cos a+\sin a\ln|\sin(x-a)|+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem tests the application of compound-angle identities in integration.
JEE Main: Recognising how to transform a trigonometric ratio into a sum of standard functions is an important problem-solving technique.
JEE Advanced: The problem reinforces the strategic use of trigonometric identities to simplify integrands before integration.
CUET and other entrance examinations: Familiarity with the standard integral of \(\cot x\) helps solve similar trigonometric integration problems efficiently.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

When the arguments of two trigonometric functions differ by a constant, consider using compound-angle identities.
The identity \(\sin x=\sin(x-a)\cos a+\cos(x-a)\sin a\) is the key to simplifying this integral.
The standard result \(\int\cot u\,du=\ln|\sin u|+C\) is useful in many trigonometric integration problems.
Always treat \(a\) as a constant while integrating with respect to \(x\).
The final expression is valid on intervals where \(\sin(x-a)\ne0\).

📘 Concept & Theory

Theory and Concept

›

This integral can be simplified by using the fundamental relationship between logarithms and exponential functions:

\[ e^{\log x}=x \]

For \(x>0\), the laws of exponents give

\[ e^{n\log x}=(e^{\log x})^n=x^n \]

Thus, the exponential expressions can be converted into powers of \(x\). The resulting rational expression can then be simplified by factorisation before applying the power rule of integration.

The standard integration formula used is

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

🗺️ Solution Roadmap

Step-by-step Plan

›

Convert each exponential-logarithmic expression into a power of \(x\).
Substitute the simplified expressions into the given integral.
Factorise the numerator and denominator.
Cancel the common factors and simplify the integrand.
Integrate the resulting polynomial using the power rule.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 8 steps

Apply the logarithmic-exponential identity
We know that
\[e^{\log x}=x\]
Therefore,
\[e^{5\log x}=x^5\]

\[e^{4\log x}=x^4\]

\[e^{3\log x}=x^3\]

\[e^{2\log x}=x^2\]
Substitute these expressions into the integral
\[I=\int\frac{x^5-x^4}{x^3-x^2}\,dx\]
Factorise the numerator and denominator
Taking the common factors,
\[x^5-x^4=x^4(x-1)\]

\[x^3-x^2=x^2(x-1)\]
Hence,
\[I=\int\frac{x^4(x-1)}{x^2(x-1)}\,dx\]
For \(x>0\) and \(x\ne1\), cancelling the common factors gives
\[I=\int x^2\,dx\]
Integrate using the power rule
Using
\[\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\]
we obtain
\[ \begin{aligned} I&=\frac{x^{2+1}}{2+1}+C\\ &=\frac{x^3}{3}+C \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{e^{5\log x}-e^{4\log x}}{e^{3\log x}-e^{2\log x}}\,dx=\frac{x^3}{3}+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem tests the use of logarithmic and exponential identities to simplify an integrand before applying basic integration formulas.
JEE Main: Recognising that \(e^{n\log x}=x^n\) can reduce a complicated-looking expression to a simple algebraic function.
JEE Advanced: The problem reinforces the importance of algebraic factorisation and identifying common factors before integration.
CUET and other entrance examinations: Such expressions test fluency in logarithmic identities, exponent laws, and elementary integration.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

For \(x>0\), the identity \(e^{n\log x}=x^n\) converts exponential-logarithmic expressions into algebraic powers.
Factorising both the numerator and denominator often reveals common factors that simplify an integral.
After simplification, apply the power rule directly whenever the resulting exponent is not \(-1\).
Always check the original domain before cancelling common factors.

📘 Concept & Theory

Theory and Concept

›

This integral contains a trigonometric function in the numerator and an expression involving its corresponding trigonometric function inside a square root in the denominator.

The numerator contains \(\cos x\), which is the derivative of \(\sin x\). Therefore, the natural substitution is

\[ t=\sin x \]

Consequently,

\[ dt=\cos x\,dx \]

This converts the trigonometric integral into an algebraic integral of the standard form

\[ \int\frac{dt}{\sqrt{a^2-t^2}} =\sin^{-1}\left(\frac{t}{a}\right)+C \]

where \(a\) is a positive constant.

🗺️ Solution Roadmap

Step-by-step Plan

›

Choose \(t=\sin x\) as the substitution.
Replace \(\cos x\,dx\) with \(dt\).
Convert the integral into the standard inverse sine form.
Apply the standard integration formula.
Substitute back \(t=\sin x\)

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Apply substitution
Let
\[t=\sin x\]
Differentiating both sides with respect to \(x\),
\[\frac{dt}{dx}=\cos x\]
Therefore,
\[dt=\cos x\,dx\]
Transform the given integral
The integral is
\[I=\int\frac{\cos x}{\sqrt{4-\sin^2x}}\,dx\]
Substituting \(t=\sin x\) and \(\cos x\,dx=dt\),
\[I=\int\frac{dt}{\sqrt{4-t^2}}\]
Apply the standard integration formula
We know that
\[\int\frac{dt}{\sqrt{a^2-t^2}}=\sin^{-1}\left(\frac{t}{a}\right)+C\]
Comparing the denominator \(\sqrt{4-t^2}\) with \(\sqrt{a^2-t^2}\),
\[a^2=4\]

\[a=2\]
Therefore,
\[I=\sin^{-1}\left(\frac{t}{2}\right)+C\]
Substitute back \(t=\sin x\)
Replacing \(t\) with \(\sin x\), we obtain
\[I=\sin^{-1}\left(\frac{\sin x}{2}\right)+C\]
\[ \boxed{ \int\frac{\cos x}{\sqrt{4-\sin^2x}}\,dx =\sin^{-1}\left(\frac{\sin x}{2}\right)+C } \]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{\cos x}{\sqrt{4-\sin^2x}}\,dx=\sin^{-1}\left(\frac{\sin x}{2}\right)+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem tests the use of substitution and the standard inverse trigonometric integration formula.
JEE Main: Recognising the relationship between the numerator and the derivative of the expression inside the square root is an important integration technique.
JEE Advanced: The problem reinforces substitution, standard integral forms, and the chain rule used in verification.
CUET and other entrance examinations: Familiarity with the form \(\int\frac{dt}{\sqrt{a^2-t^2}}\) helps solve similar trigonometric and algebraic integrals efficiently.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

When the numerator contains the derivative of the expression inside the denominator, substitution is often the most efficient approach.
The substitution \(t=\sin x\) converts the given integral into a standard algebraic form.
Remember the standard result \(\int\frac{dt}{\sqrt{a^2-t^2}}=\sin^{-1}\left(\frac{t}{a}\right)+C\).
Carefully simplify the square root when applying the inverse sine formula.

📘 Concept & Theory

Theory and Concept

›

This integral involves powers of sine and cosine in both the numerator and denominator. The main idea is to use algebraic factorisation and fundamental trigonometric identities to simplify the integrand before integrating.

The important identities are:

\[ a^4-b^4=(a^2-b^2)(a^2+b^2) \]

\[ \sin^2x+\cos^2x=1 \]

We also use the double-angle identity:

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

After simplification, the integral reduces to a standard trigonometric integral.

🗺️ Solution Roadmap

Step-by-step Plan

›

Factorise the numerator as a difference of fourth powers.
Use \(\sin^2x+\cos^2x=1\) to simplify the numerator.
Rewrite the denominator using the same trigonometric identity.
Cancel the common factor and simplify the integrand.
Integrate the resulting trigonometric expression.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 15 steps

Factorise the numerator
We have
\[\sin^8x-\cos^8x=(\sin^4x)^2-(\cos^4x)^2\]
Using the difference of two squares,
\[ \begin{aligned} \sin^8x-\cos^8x &=(\sin^4x-\cos^4x)(\sin^4x+\cos^4x)\\ &=(\sin^2x-\cos^2x) (\sin^2x+\cos^2x) (\sin^4x+\cos^4x) \end{aligned} \]
Since
\[\sin^2x+\cos^2x=1\]
we get
\[\sin^8x-\cos^8x=(\sin^2x-\cos^2x)(\sin^4x+\cos^4x)\]
Simplify the denominator
The denominator is
\[1-2\sin^2x\cos^2x\]
Using \(\sin^2x+\cos^2x=1\),
\[ \begin{aligned} 1-2\sin^2x\cos^2x &=(\sin^2x+\cos^2x)^2 -2\sin^2x\cos^2x\\ &=\sin^4x+2\sin^2x\cos^2x+\cos^4x\\ &\quad-2\sin^2x\cos^2x\\ &=\sin^4x+\cos^4x \end{aligned} \]
Therefore,
\[1-2\sin^2x\cos^2x=\sin^4x+\cos^4x\]
Substitute the simplified expressions
The given integral becomes
\[I=\int\frac{(\sin^2x-\cos^2x)(\sin^4x+\cos^4x)}{\sin^4x+\cos^4x}\,dx\]
Cancelling the common factor,
\[I=\int(\sin^2x-\cos^2x)\,dx\]
Using the double-angle identity
\[\cos 2x=\cos^2x-\sin^2x\]
we obtain
\[I=-\int\cos 2x\,dx\]
Integrate using substitution
Let
\[t=2x\]
Then
\[dt=2\,dx\]

\[dx=\frac12dt\]
Therefore,
\[ \begin{aligned} I&=-\frac12\int\cos t\,dt\\ &=-\frac12\sin t+C\\ &=-\frac12\sin 2x+C \end{aligned} \]
\[ \boxed{ \int\frac{\sin^8x-\cos^8x} {1-2\sin^2x\cos^2x}\,dx =-\frac12\sin 2x+C } \]
Equivalently, using \(\sin 2x=2\sin x\cos x\),
\[\boxed{I=-\sin x\cos x+C}\]

💡 Answer

Final Answer

›

\[ \boxed{ \int\frac{\sin^8x-\cos^8x} {1-2\sin^2x\cos^2x}\,dx =-\frac12\sin 2x+C } \]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem tests the application of algebraic factorisation and fundamental trigonometric identities in integration.
JEE Main: Recognising equivalent forms of the denominator and simplifying complicated trigonometric expressions are important problem-solving skills.
JEE Advanced: The problem demonstrates how a high-power trigonometric expression can be reduced through systematic factorisation rather than direct integration.
CUET and other entrance examinations: Knowledge of difference-of-squares identities and double-angle formulas helps simplify similar integrals efficiently.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Factorisation is often more effective than expanding high powers of trigonometric functions.
The identity \(1-2\sin^2x\cos^2x=\sin^4x+\cos^4x\) is the key to simplifying the denominator.
Use \(\sin^2x+\cos^2x=1\) to reduce expressions involving powers of sine and cosine.
The identity \(\sin^2x-\cos^2x=-\cos 2x\) converts the simplified integrand into a standard form.
Look for common factors in the numerator and denominator before attempting integration.

📘 Concept & Theory

Theory and Concept

›

This integral involves the product of two cosine functions whose arguments differ by a constant. We use the trigonometric identity:

\[ \tan A-\tan B =\frac{\sin(A-B)}{\cos A\cos B} \]

Rearranging, we obtain

\[ \frac{1}{\cos A\cos B} =\frac{\tan A-\tan B}{\sin(A-B)} \]

We also use the standard integration formula:

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

Since \(a\) and \(b\) are constants, this identity allows us to express the given integral as a difference of two standard logarithmic integrals.

🗺️ Solution Roadmap

Step-by-step Plan

›

Apply the tangent difference identity to the product of the cosine functions.
Substitute \(A=x+a\) and \(B=x+b\).
Simplify the difference \(A-B\), which is a constant.
Separate the resulting integral into two tangent integrals.
Apply the standard logarithmic integration formula.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 13 steps

Recall the tangent difference identity
We know that
\[\tan A-\tan B=\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}\]
Taking the LCM,
\[ \begin{aligned} \tan A-\tan B &=\frac{\sin A\cos B-\cos A\sin B} {\cos A\cos B}\\ &=\frac{\sin(A-B)}{\cos A\cos B} \end{aligned} \]
Therefore,
\[\frac{1}{\cos A\cos B}=\frac{\tan A-\tan B}{\sin(A-B)}\]
Substitute the given arguments
Taking
\[A=x+a,\quad B=x+b\]
we obtain
\[A-B=(x+a)-(x+b)=a-b\]
Thus,
\[\frac{1}{\cos(x+a)\cos(x+b)}=\frac{\tan(x+a)-\tan(x+b)}{\sin(a-b)}\]
Substitute this identity into the integral
\[I=\frac{1}{\sin(a-b)}\int[\tan(x+a)-\tan(x+b)]\,dx\]
Since \(a\) and \(b\) are constants,
\[I=\frac{1}{\sin(a-b)}\left[\int\tan(x+a)\,dx-\int\tan(x+b)\,dx\right]\]
Integrate using the standard formula
Recall that
\[\int\tan u\,du=-\ln|\cos u|+C\]
For the first integral, putting \(u=x+a\), we get
\[\int\tan(x+a)\,dx=-\ln|\cos(x+a)|\]
Similarly, for the second integral,
\[\int\tan(x+b)\,dx=-\ln|\cos(x+b)|\]
Therefore,
\[\begin{aligned}I&=\frac{1}{\sin(a-b)}\left[-\ln|\cos(x+a)|+\ln|\cos(x+b)|\right]+C\\ &=\frac{1}{\sin(a-b)}\ln\left|\frac{\cos(x+b)}{\cos(x+a)}\right|+C\end{aligned}\]
For \(\sin(a-b)\ne0\), the required integral is
\[\boxed{\int\frac{dx}{\cos(x+a)\cos(x+b)}=\frac{1}{\sin(a-b)}\ln\left|\frac{\cos(x+b)}{\cos(x+a)}\right|+C}\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{dx}{\cos(x+a)\cos(x+b)}=\frac{1}{\sin(a-b)}\ln\left|\frac{\cos(x+b)}{\cos(x+a)}\right|+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem tests the application of trigonometric identities in simplifying and evaluating integrals.
JEE Main: Recognising the difference of two tangent functions as a useful transformation helps simplify products of cosine functions.
JEE Advanced: The problem develops the ability to identify suitable trigonometric transformations and handle special parameter values.
CUET and other entrance examinations: The standard logarithmic integral of the tangent function is frequently useful in solving trigonometric integration problems.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

When the arguments of two cosine functions differ by a constant, consider using the tangent difference identity.
The identity \(\tan A-\tan B=\frac{\sin(A-B)}{\cos A\cos B}\) is the central step in this problem.
Remember that \(\int\tan x\,dx=-\ln|\cos x|+C\).
Pay attention to the parameter condition \(\sin(a-b)\ne0\) before dividing by \(\sin(a-b)\).
When \(\sin(a-b)=0\), simplify the original integrand separately instead of using the general logarithmic formula.

📘 Concept & Theory

Theory and Concept

›

This integral contains a power of \(x\) in the numerator and a square root involving \(x^8\) in the denominator. The numerator suggests the substitution \(t=x^4\), since its derivative contains \(x^3\).

We use the standard integral formula:

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

The objective is to transform the given integral into this standard inverse sine form.

🗺️ Solution Roadmap

Step-by-step Plan

›

Choose \(t=x^4\) as the substitution.
Differentiate to express \(x^3\,dx\) in terms of \(dt\).
Rewrite the square root in terms of \(t\).
Apply the standard inverse sine integration formula.
Substitute back \(t=x^4\) and verify the result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Apply substitution
Let
\[t=x^4\]
Differentiating both sides with respect to \(x\),
\[dt=4x^3\,dx\]
Therefore,
\[x^3\,dx=\frac14\,dt\]
Transform the given integral
Since \(t=x^4\),
\[x^8=(x^4)^2=t^2\]
Substituting these expressions into the integral,
\[ \begin{aligned} I &=\int\frac{x^3}{\sqrt{1-x^8}}\,dx\\ &=\frac14\int\frac{dt}{\sqrt{1-t^2}} \end{aligned} \]
Apply the standard integration formula
We know that
\[\int\frac{dt}{\sqrt{1-t^2}}=\sin^{-1}t+C\]
Hence,
\[I=\frac14\sin^{-1}t+C\]
Substitute back \(t=x^4\)
Replacing \(t\) with \(x^4\), we obtain
\[I=\frac14\sin^{-1}(x^4)+C\]
\[\boxed{\int\frac{x^3}{\sqrt{1-x^8}}\,dx=\frac14\sin^{-1}(x^4)+C}\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{x^3}{\sqrt{1-x^8}}\,dx=\frac14\sin^{-1}(x^4)+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem tests substitution and the application of a standard inverse trigonometric integration formula.
JEE Main: Identifying the appropriate substitution from the relationship between the numerator and the expression inside the square root is an important integration skill.
JEE Advanced: The problem reinforces the use of composite-function differentiation in reverse, which is the underlying principle of substitution.
CUET and other entrance examinations: Recognising the standard form \(\int\frac{dt}{\sqrt{1-t^2}}\) helps simplify apparently complicated algebraic integrals.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

When the numerator contains a power that is closely related to the expression inside the square root, look for a suitable substitution.
The substitution \(t=x^4\) transforms \(x^3\,dx\) into \(\frac14dt\).
The expression \(1-x^8\) becomes \(1-t^2\), giving the standard inverse sine integral.
Remember the standard result \(\int\frac{dt}{\sqrt{1-t^2}}=\sin^{-1}t+C\).
Always verify an indefinite integral by differentiation.

📘 Concept & Theory

Theory and Concept

›

The integrand contains \(e^x\) in the numerator as well as in the denominator. Since the derivative of \(e^x\) is also \(e^x\), the natural substitution is

\[ t=e^x \]

Then,

\[ dt=e^x\,dx \]

The integral becomes a rational algebraic function that can be evaluated using partial fraction decomposition.

We use the standard results:

\[ \int\frac{dt}{t+a}=\ln|t+a|+C \]

\[ \frac{1}{(t+1)(t+2)} =\frac{1}{t+1}-\frac{1}{t+2} \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Substitute \(t=e^x\) and express \(e^x\,dx\) as \(dt\).
Convert the given integral into a rational function of \(t\).
Decompose the rational function into partial fractions.
Integrate the resulting logarithmic terms.
Substitute back \(t=e^x\).
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 15 steps

Apply substitution
Let
\[t=e^x\]
Differentiating both sides with respect to \(x\),
\[dt=e^x\,dx\]
Therefore,
\[e^x\,dx=dt\]
Transform the given integral
Substituting \(t=e^x\) and \(e^x\,dx=dt\), we get
\[ \begin{aligned} I &=\int\frac{e^x}{(1+e^x)(2+e^x)}\,dx\\ &=\int\frac{dt}{(1+t)(2+t)} \end{aligned} \]
Apply partial fraction decomposition
Let
\[\frac{1}{(t+1)(t+2)}=\frac{A}{t+1}+\frac{B}{t+2}\]
Taking the LCM,
\[1=A(t+2)+B(t+1)\]
Expanding,
\[1=(A+B)t+(2A+B)\]
Comparing coefficients,
\[A+B=0 \tag{1}\]

\[2A+B=1 \tag{2}\]
Subtracting equation (1) from equation (2),
\[A=1\]
Using equation (1),
\[B=-1\]
Therefore,
\[\frac{1}{(t+1)(t+2)}=\frac{1}{t+1}-\frac{1}{t+2}\]
Integrate term by term
Substituting the partial fractions,
\[I=\int\left(\frac{1}{t+1}-\frac{1}{t+2}\right)dt\]
Integrating,
\[ \begin{aligned} I &=\ln|t+1|-\ln|t+2|+C\\ &=\ln\left|\frac{t+1}{t+2}\right|+C \end{aligned} \]
Substitute back \(t=e^x\).
Replacing \(t\) with \(e^x\),
\[I=\ln\left|\frac{1+e^x}{2+e^x}\right|+C\]
Since \(e^x>0\), both \(1+e^x\) and \(2+e^x\) are positive. Therefore, the absolute value signs can be omitted.
\[\boxed{\int\frac{e^x}{(1+e^x)(2+e^x)}\,dx=\ln\left(\frac{1+e^x}{2+e^x}\right)+C}\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{e^x}{(1+e^x)(2+e^x)}\,dx=\ln\left(\frac{1+e^x}{2+e^x}\right)+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem tests substitution and partial fraction decomposition in the integration of exponential functions.
JEE Main: Recognising \(e^x\,dx\) as the differential of \(e^x\) helps transform exponential integrals into rational algebraic forms.
JEE Advanced: The problem reinforces the combination of substitution, coefficient comparison, and logarithmic integration.
CUET and other entrance examinations: Familiarity with exponential substitutions and partial fractions helps solve similar integrals accurately and efficiently.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

When an exponential function and its derivative appear together, consider the substitution \(t=e^x\).
After substitution, the integral becomes a rational function that may be suitable for partial fraction decomposition.
The decomposition \(\frac{1}{(t+1)(t+2)}=\frac{1}{t+1}-\frac{1}{t+2}\) simplifies the integration.
The difference of logarithms can be combined into a single logarithm of a quotient.
Always differentiate the final expression to verify the result.

📘 Concept & Theory

Theory and Concept

›

The given integrand is a rational algebraic function whose denominator is the product of two distinct irreducible quadratic factors.

For such expressions, we use partial fraction decomposition of the form

\[ \frac{1}{(x^2+1)(x^2+4)} =\frac{Ax+B}{x^2+1} +\frac{Cx+D}{x^2+4} \]

However, since the numerator is an even function of \(x\), a simpler decomposition is sufficient:

\[ \frac{1}{(x^2+1)(x^2+4)} =\frac{A}{x^2+1}+\frac{B}{x^2+4} \]

We then use the standard integral formula

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

🗺️ Solution Roadmap

Step-by-step Plan

›

Express the rational function as a sum of two partial fractions.
Determine the unknown constants by comparing coefficients.
Split the integral into two standard inverse tangent integrals.
Apply the standard formula for \(\int\frac{dx}{x^2+a^2}\).
Combine the results and verify by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Apply partial fraction decomposition
Let
\[\frac{1}{(x^2+1)(x^2+4)}=\frac{A}{x^2+1}+\frac{B}{x^2+4}\]
Taking the LCM,
\[1=A(x^2+4)+B(x^2+1)\]
Expanding,
\[1=Ax^2+4A+Bx^2+B\]

\[1=(A+B)x^2+(4A+B)\]
Comparing coefficients of corresponding powers of \(x\),
\[A+B=0 \tag{1}\]

\[4A+B=1 \tag{2}\]
Subtracting equation (1) from equation (2),
\[3A=1\]

\[\boxed{A=\frac13}\]
Substituting \(A=\frac13\) into equation (1),
\[\frac13+B=0\]

\[\boxed{B=-\frac13}\]
Therefore, the required partial fraction decomposition is
\[\boxed{\frac{1}{(x^2+1)(x^2+4)}=\frac13\left(\frac{1}{x^2+1}-\frac{1}{x^2+4}\right)}\]
Substitute the decomposition into the integral
\[\begin{aligned}I&=\frac13\int\left(\frac{1}{x^2+1}-\frac{1}{x^2+4}\right)dx\\ &=\frac13\left[\int\frac{dx}{x^2+1}-\int\frac{dx}{x^2+4}\right]\end{aligned}\]
Evaluate the first integral
Using the standard formula with \(a=1\),
\[\int\frac{dx}{x^2+1}=\tan^{-1}x\]
Evaluate the second integral
Since \(4=2^2\), applying the standard formula with \(a=2\),
\[\int\frac{dx}{x^2+4}=\frac12\tan^{-1}\left(\frac{x}{2}\right)\]
Combine the results
Substituting both results,
\[\begin{aligned}I&=\frac13\left[\tan^{-1}x-\frac12\tan^{-1}\left(\frac{x}{2}\right)\right]+C\\ &=\frac13\tan^{-1}x-\frac16\tan^{-1}\left(\frac{x}{2}\right)+C\end{aligned}\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{dx}{(x^2+1)(x^2+4)}=\frac13\tan^{-1}x-\frac16\tan^{-1}\left(\frac{x}{2}\right)+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem tests partial fraction decomposition involving irreducible quadratic factors and the application of standard inverse tangent integrals.
JEE Main: Recognising the appropriate partial fraction form and applying inverse tangent formulas are important techniques for rational function integration.
JEE Advanced: The problem reinforces algebraic decomposition and the efficient evaluation of rational functions with quadratic denominators.
CUET and other entrance examinations: Such integrals frequently test the ability to combine partial fractions with standard integration formulas.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For a product of distinct irreducible quadratic factors with constant numerators, use partial fractions with constant numerators.
Comparing coefficients gives \(A=\frac13\) and \(B=-\frac13\).
Remember the standard result \(\int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C\).
Keep the negative sign associated with the second partial fraction throughout the integration.
Differentiate the final expression to confirm the correctness of the result.

📘 Concept & Theory

Theory and Concept

›

This problem involves two important concepts:

Concept 1: Exponential and logarithmic identity

For a positive real number \(a\),

\[ e^{\log a}=a \]

Therefore, wherever \(\sin x>0\),

\[ e^{\log\sin x}=\sin x \]

Concept 2: Trigonometric substitution

When an integral contains a product of an odd power of cosine and sine, it is often convenient to separate one factor of cosine and use the substitution

\[ u=\sin x,\qquad du=\cos x\,dx \]

Alternatively, we can use \(u=\cos x\) after expressing the integrand in terms of \(\sin x\) and \(\cos x\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Simplify the exponential-logarithmic expression.
Express \(\cos^3x\) as \(\cos^2x\cdot\cos x\).
Use the identity \(\cos^2x=1-\sin^2x\).
Substitute \(u=\sin x\) and integrate the resulting polynomial.
Differentiate the final expression to verify the result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 8 steps

Given —
\[I=\int \cos^3x\,e^{\log\sin x}\,dx\]
Using the identity \(e^{\log a}=a\), we obtain
\[I=\int\cos^3x\sin x\,dx\]
Rewrite the integrand
Since
\[\cos^3x=\cos^2x\cdot\cos x\]
and
\[\cos^2x=1-\sin^2x,\]
we get
\[\begin{aligned}I&=\int(1-\sin^2x)\sin x\cos x\,dx\\ &=\int(1-\sin^2x)\sin x\cos x\,dx\end{aligned}\]
Apply substitution
Let
\[u=\sin x\]
Differentiating with respect to \(x\),
\[du=\cos x\,dx\]
Substituting \(u=\sin x\) and \(du=\cos x\,dx\),
\[\begin{aligned}I&=\int(1-u^2)u\,du\\ &=\int(u-u^3)\,du\end{aligned}\]
Integrate term by term
\[\begin{aligned}I&=\int u\,du-\int u^3\,du\\ &=\frac{u^2}{2}-\frac{u^4}{4}+C\end{aligned}\]
Substitute back \(u=\sin x\)
\[I=\frac{\sin^2x}{2}-\frac{\sin^4x}{4}+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\cos^3x\,e^{\log\sin x}\,dx=\frac{\sin^2x}{2}-\frac{\sin^4x}{4}+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem tests the simplification of exponential-logarithmic expressions and the use of trigonometric identities in integration.
JEE Main: Recognising the substitution \(u=\sin x\) and converting the integrand into a polynomial are useful techniques for evaluating trigonometric integrals.
JEE Advanced: The problem reinforces the strategic selection of substitutions after simplifying the given expression.
CUET and other entrance examinations: Questions of this type assess familiarity with logarithmic identities, trigonometric identities and elementary integration formulas.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Use \(e^{\log a}=a\) for \(a>0\) to simplify exponential-logarithmic expressions.
For an odd power of cosine, separate one factor of \(\cos x\) and use \(u=\sin x\).
Apply \(\cos^2x=1-\sin^2x\) to convert the remaining trigonometric expression into a polynomial.
Always check the domain of logarithmic expressions before simplifying them.
Verification by differentiation confirms the correctness of the antiderivative.

📘 Concept & Theory

Theory and Concept

›

This problem involves two important concepts.

Concept 1: Exponential-logarithmic identity

For \(x>0\),

\[ e^{\log x}=x \]

Concept 2: Substitution method

When an integral contains a rational expression involving \(x^4+1\) and a factor of \(x\) in the numerator, the substitution

\[ u=x^2,\quad du=2x\,dx \]

reduces the integral to a standard form involving \(u^2+1\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Simplify the exponential-logarithmic expression.
Rewrite the integral as a rational function.
Use the substitution \(u=x^2\).
Apply the standard inverse tangent integration formula.
Substitute back and verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Simplify the given expression
We have
\[I=\int e^{\log x}(x^4+1)^{-1}\,dx\]
Using \(e^{\log x}=x\),
\[I=\int\frac{x}{x^4+1}\,dx\]
Apply substitution
Let
\[u=x^2\]
Differentiating both sides,
\[du=2x\,dx\]
Therefore,
\[x\,dx=\frac12\,du\]
Also,
\[x^4+1=(x^2)^2+1=u^2+1\]
Substituting these expressions into the integral,
\[\begin{aligned}I&=\int\frac{x}{x^4+1}\,dx\\ &=\frac12\int\frac{du}{u^2+1}\end{aligned}\]
Apply the standard integration formula
Recall that
\[\int\frac{du}{1+u^2}=\tan^{-1}u+C\]
Hence,
\[I=\frac12\tan^{-1}u+C\]
Substitute back \(u=x^2\)
\[I=\frac12\tan^{-1}(x^2)+C\]
\[\boxed{\int e^{\log x}(x^4+1)^{-1}\,dx=\frac12\tan^{-1}(x^2)+C}\]

💡 Answer

Final Answer

›

\[\boxed{\int e^{\log x}(x^4+1)^{-1}\,dx=\frac12\tan^{-1}(x^2)+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem tests the simplification of exponential-logarithmic expressions and the application of substitution in integration.
JEE Main: Recognising the substitution \(u=x^2\) converts a quartic rational expression into the standard inverse tangent form.
JEE Advanced: The problem reinforces the ability to identify hidden standard integral forms through suitable algebraic substitutions.
CUET and other entrance examinations: This is a useful application of logarithmic identities, the chain rule and standard integration formulas.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For \(x>0\), simplify \(e^{\log x}\) directly to \(x\).
Recognise that the numerator \(x\) is the derivative of \(x^2\), apart from a constant factor.
The substitution \(u=x^2\) transforms \(x^4+1\) into \(u^2+1\).
Remember the standard result \(\int\frac{du}{1+u^2}=\tan^{-1}u+C\).
Verification by differentiation confirms the coefficient \(\frac12\) in the final answer.

📘 Concept & Theory

Theory and Concept

›

This problem is based on the reverse chain rule of differentiation and the substitution method of integration.

Recall the standard integration formula:

\[ \int u^n\,du=\frac{u^{n+1}}{n+1}+C, \qquad n\ne-1 \]

Also, by the chain rule,

\[ \frac{d}{dx}f(ax+b)=a f'(ax+b) \]

Therefore, the given integral can be evaluated by substituting

\[ u=f(ax+b) \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the inner function \(f(ax+b)\).
Substitute \(u=f(ax+b)\).
Differentiate the substitution to express \(f'(ax+b)\,dx\) in terms of \(du\).
Apply the power rule of integration.
Substitute back.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 7 steps

Given —
\[I=\int f'(ax+b)\left[f(ax+b)\right]^n\,dx\]
Apply substitution
Let
\[u=f(ax+b)\]
Differentiating both sides with respect to \(x\), using the chain rule,
\[\frac{du}{dx}=a f'(ax+b)\]
Therefore,
\[du=a f'(ax+b)\,dx\]
or
\[f'(ax+b)\,dx=\frac{du}{a}\]
Transform the integral
Substituting \(u=f(ax+b)\) and \(f'(ax+b)\,dx=\frac{du}{a}\),
\[ \begin{aligned} I &=\int f'(ax+b)\left[f(ax+b)\right]^n\,dx\\ &=\frac1a\int u^n\,du \end{aligned} \]
Apply the power rule of integration
For \(n\ne-1\),
\[ \begin{aligned} I &=\frac1a\cdot\frac{u^{n+1}}{n+1}+C\\ &=\frac{u^{n+1}}{a(n+1)}+C \end{aligned} \]
Substitute back \(u=f(ax+b)\)
\[I=\frac{\left[f(ax+b)\right]^{n+1}}{a(n+1)}+C\]
For \(a\ne0\) and \(n\ne-1\),
\[ \boxed{ \int f'(ax+b)\left[f(ax+b)\right]^n\,dx = \frac{\left[f(ax+b)\right]^{n+1}}{a(n+1)}+C } \]

💡 Answer

Final Answer

›

\[ \boxed{ \int f'(ax+b)\left[f(ax+b)\right]^n\,dx = \frac{\left[f(ax+b)\right]^{n+1}}{a(n+1)}+C } \]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem tests the application of the substitution method and the reverse chain rule in integration.
JEE Main: Recognising the derivative of a composite function is essential for quickly evaluating integrals of this form.
JEE Advanced: The problem develops a generalised approach to integrating expressions involving an arbitrary differentiable function.
CUET and other entrance examinations: The formula is useful for identifying standard integration patterns involving powers of composite functions.

📜 Note

Important Notes

›

The formula for \(n\ne-1\) requires \(a\ne0\).
For \(n=-1\), the logarithmic formula applies wherever \(f(ax+b)\ne0\).
The function \(f\) is assumed to be differentiable on the relevant interval.
If \(a=0\), the integrand becomes constant with respect to \(x\), and the integral is \(f'(b)[f(b)]^n x+C\), whenever the expression is defined.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Identify the inner function \(f(ax+b)\) before choosing a substitution.
Always account for the constant multiplier \(a\) arising from the chain rule.
For \(n\ne-1\), use the power rule of integration.
For \(n=-1\), use the logarithmic integration formula.
This is a generalised standard integral based on the reverse chain rule.

📘 Concept & Theory

Theory and Concept

›

This problem involves the following trigonometric identity and substitution technique.

Concept 1: Compound-angle identity

Using the identity

\[ \sin(x+\alpha)=\sin x\cos\alpha+\cos x\sin\alpha \]

we can factor out \(\sin x\) to obtain

\[ \sin(x+\alpha) =\sin x(\cos\alpha+\cot x\sin\alpha) \]

Concept 2: Cotangent substitution

The derivative of \(\cot x\) is

\[ \frac{d}{dx}(\cot x)=-\csc^2x =-\frac{1}{\sin^2x} \]

Thus, the substitution

\[ t=\cot x \]

is particularly useful when the integrand contains the factor \(\frac{1}{\sin^2x}\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Expand \(\sin(x+\alpha)\) using the compound-angle identity.
Factor the expression under the square root.
Rewrite the integral in terms of \(\cot x\).
Apply the substitution \(t=\cot x\).
Evaluate the resulting standard integral and substitute back.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 15 steps

Apply the compound-angle identity
We have
\[I=\int\frac{dx}{\sqrt{\sin^3x\sin(x+\alpha)}}\]
Using
\[\sin(x+\alpha)=\sin x\cos\alpha+\cos x\sin\alpha\]
we get
\[ \begin{aligned} I &=\int\frac{dx} {\sqrt{\sin^3x(\sin x\cos\alpha+\cos x\sin\alpha)}}\\ &=\int\frac{dx} {\sqrt{\sin^4x\cos\alpha+ \sin^3x\cos x\sin\alpha}} \end{aligned} \]
Factor out \(\sin^4x\) under the square root
Taking \(\sin^4x\) as a common factor,
\[ \begin{aligned} I &=\int\frac{dx} {\sqrt{\sin^4x\left(\cos\alpha+ \frac{\cos x}{\sin x}\sin\alpha\right)}}\\ &=\int\frac{dx} {\sqrt{\sin^4x(\cos\alpha+\cot x\sin\alpha)}} \end{aligned} \]
Since \(\sqrt{\sin^4x}=\sin^2x\),
\[I=\int\frac{dx}{\sin^2x\sqrt{\cos\alpha+\cot x\sin\alpha}}\]
Apply substitution
Let
\[t=\cot x\]
Differentiating with respect to \(x\),
\[dt=-\csc^2x\,dx\]
Therefore,
\[dt=-\frac{dx}{\sin^2x}\]
or
\[\frac{dx}{\sin^2x}=-dt\]
Substituting into the integral,
\[\begin{aligned}I&=-\int\frac{dt}{\sqrt{\cos\alpha+t\sin\alpha}}\end{aligned}\]
Evaluate the resulting integral
Let
\[u=\cos\alpha+t\sin\alpha\]
Differentiating
\[du=\sin\alpha\,dt\]
Hence, provided \(\sin\alpha\ne0\),
\[dt=\frac{du}{\sin\alpha}\]
Substituting,
\[ \begin{aligned} I &=-\frac{1}{\sin\alpha} \int\frac{du}{\sqrt{u}}\\ &=-\frac{1}{\sin\alpha} \int u^{-1/2}\,du\\ &=-\frac{1}{\sin\alpha} \cdot\frac{u^{1/2}}{1/2}+C\\ &=-\frac{2}{\sin\alpha}\sqrt{u}+C \end{aligned} \]
Substitute back \(u=\cos\alpha+t\sin\alpha\) and \(t=\cot x\)
\[I=-\frac{2}{\sin\alpha}\sqrt{\cos\alpha+\sin\alpha\cot x}+C\]
For \(\sin\alpha\ne0\),
\[\boxed{\int\frac{dx}{\sqrt{\sin^3x\sin(x+\alpha)}}=-\frac{2}{\sin\alpha}\sqrt{\cos\alpha+\sin\alpha\cot x}+C}\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{dx}{\sqrt{\sin^3x\sin(x+\alpha)}}=-\frac{2}{\sin\alpha}\sqrt{\cos\alpha+\sin\alpha\cot x}+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem tests the application of compound-angle identities and substitution in trigonometric integration.
JEE Main: Recognising the usefulness of the cotangent substitution is important for simplifying integrals involving powers of sine and compound angles.
JEE Advanced: The problem develops the ability to transform complicated radical expressions into standard integrals through suitable trigonometric identities.
CUET and other entrance examinations: It reinforces the relationship between trigonometric identities, derivatives and standard integration formulas.

📜 Note

Important Notes

›

The original integral is real-valued only where \(\sin^3x\sin(x+\alpha)>0\).
The derivation assumes \(\sin\alpha\ne0\); the case \(\alpha=2k\pi\) must be evaluated separately.
The integration constant may differ on separate connected intervals of the real domain.
The identity \(\sqrt{\sin^4x}=\sin^2x\), rather than \(\sin x\), is essential for obtaining the correct result.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Expand compound angles before attempting a complicated trigonometric integral.
Factor the expression under the radical to identify a suitable substitution.
Use \(t=\cot x\) when the differential contains \(\frac{dx}{\sin^2x}\).
Remember that \(\int u^{-1/2}\,du=2\sqrt{u}+C\).
Check exceptional parameter values separately when a substitution requires division by a parameter.

📘 Concept & Theory

Theory and Concept

›

The given integral contains nested square roots. Such integrals can often be simplified by choosing a substitution that eliminates the radical expression.

We use the substitution

\[ t=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \]

Squaring and rearranging gives

\[ \sqrt{x}=\frac{1-t^2}{1+t^2} \]

This converts the original integral into a rational function of \(t\), which can be evaluated using polynomial integration and the standard inverse tangent formula.

🗺️ Solution Roadmap

Step-by-step Plan

›

Introduce a substitution equal to the entire radical expression.
Express \(\sqrt{x}\) and \(x\) in terms of the new variable.
Differentiate the resulting expression for \(x\).
Transform the integral into a rational function.
Evaluate the resulting polynomial and inverse tangent integrals.
Substitute back and verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 26 steps

Apply substitution
Let
\[t=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\]
Squaring both sides,
\[t^2=\frac{1-\sqrt{x}}{1+\sqrt{x}}\]
Cross-multiplying,
\[t^2(1+\sqrt{x})=1-\sqrt{x}\]
Expanding,
\[t^2+t^2\sqrt{x}=1-\sqrt{x}\]
Collecting the terms involving \(\sqrt{x}\),
\[\sqrt{x}(1+t^2)=1-t^2\]
Therefore,
\[\boxed{\sqrt{x}=\frac{1-t^2}{1+t^2}}\]
Squaring both sides,
\[\boxed{x=\frac{(1-t^2)^2}{(1+t^2)^2}}\]
Differentiate with respect to \(t\)
We have
\[x=\left(\frac{1-t^2}{1+t^2}\right)^2\]
Using the chain rule,
\[\begin{aligned}\frac{dx}{dt}&=2\left(\frac{1-t^2}{1+t^2}\right)\frac{d}{dt}\left(\frac{1-t^2}{1+t^2}\right)\end{aligned}\]
Applying the quotient rule,
\[ \begin{aligned} \frac{d}{dt}\left(\frac{1-t^2}{1+t^2}\right) &=\frac{(1+t^2)(-2t)-(1-t^2)(2t)} {(1+t^2)^2}\\ &=\frac{-2t-2t^3-2t+2t^3} {(1+t^2)^2}\\ &=\frac{-4t}{(1+t^2)^2} \end{aligned} \]
Hence,
\[ \begin{aligned} \frac{dx}{dt} &=2\left(\frac{1-t^2}{1+t^2}\right) \left(\frac{-4t}{(1+t^2)^2}\right)\\ &=-\frac{8t(1-t^2)}{(1+t^2)^3} \end{aligned} \]
Therefore,
\[dx=-\frac{8t(1-t^2)}{(1+t^2)^3}\,dt\]
Transform the integral
Since
\[\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}=t\]
the integral becomes
\[ \begin{aligned} I &=\int t\left[-\frac{8t(1-t^2)} {(1+t^2)^3}\right]dt\\ &=-8\int\frac{t^2(1-t^2)}{(1+t^2)^3}\,dt \end{aligned} \]
Simplify the rational expression
Observe that
\[t^2(1-t^2)=t^2-t^4\]
We can express the integrand as
\[\frac{t^2-t^4}{(1+t^2)^3}=-\frac{1}{1+t^2}+\frac{3}{(1+t^2)^2}-\frac{2}{(1+t^2)^3}\]
Thus,
\[I=8\int\left[\frac{1}{1+t^2}-\frac{3}{(1+t^2)^2}+\frac{2}{(1+t^2)^3}\right]dt\]
Evaluate the standard integrals
We use the standard results
\[\int\frac{dt}{1+t^2}=\tan^{-1}t+C\]

\[\int\frac{dt}{(1+t^2)^2}=\frac{t}{2(1+t^2)}+\frac12\tan^{-1}t+C\]

\[\int\frac{dt}{(1+t^2)^3}=\frac{t}{4(1+t^2)^2}+\frac{3t}{8(1+t^2)}+\frac38\tan^{-1}t+C\]
Substituting these results,
\[ \begin{aligned} I=8\Bigg[ &\tan^{-1}t -3\left\{\frac{t}{2(1+t^2)} +\frac12\tan^{-1}t\right\}\\ &+2\left\{\frac{t}{4(1+t^2)^2} +\frac{3t}{8(1+t^2)} +\frac38\tan^{-1}t\right\} \Bigg]+C \end{aligned} \]
Collecting the inverse tangent terms,
\[1-\frac32+\frac34=\frac14\]
Collecting the rational terms,
\[-\frac{3t}{2(1+t^2)}+\frac{t}{2(1+t^2)^2}+\frac{3t}{4(1+t^2)}\]
Therefore,
\[\begin{aligned}I&=8\left[\frac14\tan^{-1}t-\frac{3t}{4(1+t^2)}+\frac{t}{2(1+t^2)^2}\right]+C\\ &=2\tan^{-1}t-\frac{6t}{1+t^2}+\frac{4t}{(1+t^2)^2}+C\end{aligned}\]
Substitute back in terms of \(x\)
Recall that
\[t=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\]
Also,
\[1+t^2=1+\frac{1-\sqrt{x}}{1+\sqrt{x}}=\frac{2}{1+\sqrt{x}}\]
Consequently,
\[\frac{t}{1+t^2}=\frac12\sqrt{1-x}\]
and
\[\frac{t}{(1+t^2)^2}=\frac{1+\sqrt{x}}{4}\sqrt{1-x}\]
Substituting these expressions,
\[ \begin{aligned} I &=2\tan^{-1}\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\\ &\quad-3\sqrt{1-x} +(1+\sqrt{x})\sqrt{1-x}+C\\ &=2\tan^{-1}\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\\ &\quad+(\sqrt{x}-2)\sqrt{1-x}+C \end{aligned} \]

💡 Answer

Final Answer

›

\[ \boxed{ \begin{aligned} \int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,dx ={}&(\sqrt{x}-2)\sqrt{1-x}\\ &+2\tan^{-1}\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C \end{aligned} } \]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem tests the use of algebraic substitutions to simplify nested radicals and evaluate integrals.
JEE Main: Recognising a substitution that converts a radical expression into a rational function is an important integration technique.
JEE Advanced: The problem develops proficiency in rational-function integration and the evaluation of standard integrals involving powers of \(1+t^2\).
CUET and other entrance examinations: It reinforces substitution, algebraic manipulation and inverse tangent integration.

📜 Note

Important Notes

›

For the original integrand to be real-valued, the natural domain is \(0
The substitution \(t=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\) is valid on \(0
The endpoint values can be handled by continuity, although the derivation is performed on the open interval.
The integration constant is understood on the connected interval of the domain.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For nested radical expressions, consider substituting the entire radical as a new variable.
Express the original variable in terms of the new variable before differentiating.
Careful simplification can transform a complicated radical integral into a rational integral.
Standard integrals involving \(\frac{1}{(1+t^2)^2}\) and \(\frac{1}{(1+t^2)^3}\) are useful in advanced integration.
Always verify a complicated antiderivative by differentiation.

📘 Concept & Theory

Theory and Concept

›

This problem involves two important techniques:

Concept 1: Double-angle identities

Recall that

\[ \sin 2x=2\sin x\cos x \]

\[ 1+\cos 2x=2\cos^2x \]

These identities simplify the trigonometric fraction.

Concept 2: Exponential substitution and partial fractions

The substitution

\[ t=e^x,\qquad dt=e^x\,dx \]

is useful when the integral contains \(e^x\,dx\). It converts the resulting expression into a rational function of \(t\), which can be evaluated using partial fractions.

🗺️ Solution Roadmap

Step-by-step Plan

›

Expand the trigonometric fraction using double-angle identities.
Express the simplified integrand in terms of \(\tan x\).
Use the substitution \(t=e^x\).
Apply partial fraction decomposition.
Integrate the resulting rational function.
Substitute back and verify the result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 13 steps

Simplify the trigonometric fraction
We have
\[I=\int\frac{2+\sin 2x}{1+\cos 2x}e^x\,dx\]
Using the identities
\[\sin 2x=2\sin x\cos x\]
and
\[1+\cos 2x=2\cos^2x,\]
we obtain
\[ \begin{aligned} \frac{2+\sin 2x}{1+\cos 2x} &=\frac{2+2\sin x\cos x}{2\cos^2x}\\ &=\frac{1+\sin x\cos x}{\cos^2x}\\ &=\sec^2x+\tan x \end{aligned} \]
Therefore,
\[I=\int e^x(\sec^2x+\tan x)\,dx\]
Rewrite the trigonometric expression
Since
\[\sec^2x=1+\tan^2x,\]
we can write
\[I=\int e^x(1+\tan^2x+\tan x)\,dx\]
Apply exponential substitution
Let
\[t=e^x\]
Then
\[dt=e^x\,dx\]
Also, \(x=\log t\). Thus, the integral becomes
\[I=\int\left[1+\tan^2(\log t)+\tan(\log t)\right]dt\]
This substitution does not produce a rational function of \(t\). Therefore, partial fraction decomposition is not applicable here.
Use a suitable direct observation
Observe that
\[\frac{d}{dx}(e^x\tan x)=e^x\tan x+e^x\sec^2x\]
Hence,
\[\frac{d}{dx}(e^x\tan x)=e^x(\tan x+\sec^2x)\]
This is exactly the simplified integrand
Therefore,
\[I=e^x\tan x+C\]
\[\boxed{\int\frac{2+\sin 2x}{1+\cos 2x}e^x\,dx=e^x\tan x+C}\]

💡 Answer

Final Answer

›

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

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem tests the use of double-angle identities, trigonometric simplification and the product rule in reverse.
JEE Main: Recognising the derivative of \(e^x\tan x\) helps evaluate the integral directly without unnecessary algebra.
JEE Advanced: The problem reinforces the importance of identifying hidden derivatives before selecting a substitution.
CUET and other entrance examinations: It tests the combined application of trigonometric identities and differentiation-based integration techniques.

📜 Note

Important Notes

›

The original denominator is

\[ 1+\cos 2x=2\cos^2x \]

Therefore, the integrand is defined wherever

\[ \cos x\ne0 \]

The antiderivative is valid separately on each interval where the original integrand is defined.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Use double-angle identities to simplify complicated trigonometric fractions.
Recognise that \(\frac{2+\sin 2x}{1+\cos 2x}=\sec^2x+\tan x\).
The product rule gives \(\frac{d}{dx}(e^x\tan x)=e^x(\sec^2x+\tan x)\).
Identifying a hidden derivative is often more efficient than attempting a lengthy substitution.
Always check the domain restrictions imposed by the original denominator.

📘 Concept & Theory

Theory and Concept

›

The given integrand is a proper rational function with a repeated linear factor and a distinct linear factor in its denominator.

For a denominator containing a repeated linear factor \((x+a)^2\) and another distinct linear factor, the partial fraction decomposition is written as

\[ \frac{P(x)}{(x+a)^2(x+b)} = \frac{A}{x+a} +\frac{B}{(x+a)^2} +\frac{C}{x+b} \]

After determining the constants, we use the standard integration formulas:

\[ \int\frac{dx}{x+a}=\log|x+a|+C \]

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

🗺️ Solution Roadmap

Step-by-step Plan

›

Express the rational function in partial fractions.
Multiply by the common denominator and compare coefficients.
Determine the unknown constants.
Integrate each partial fraction separately.
Combine the logarithmic and algebraic terms.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 17 steps

Apply partial fraction decomposition
Let
\[\frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+2}\]
Taking the LCM,
\[ \begin{aligned} x^2+x+1 ={}&A(x+1)(x+2)\\ &+B(x+2)+C(x+1)^2 \end{aligned} \]
Expand the expression
Expanding each term,
\[ \begin{aligned} x^2+x+1 ={}&A(x^2+3x+2)\\ &+Bx+2B+C(x^2+2x+1) \end{aligned} \]
Collecting like terms,
\[ \begin{aligned} x^2+x+1 ={}&(A+C)x^2\\ &+(3A+B+2C)x\\ &+(2A+2B+C) \end{aligned} \]
Compare coefficients
Comparing the coefficients of corresponding powers of \(x\), we obtain
\[A+C=1 \tag{1}\]

\[3A+B+2C=1 \tag{2}\]

\[2A+2B+C=1 \tag{3}\]
From equation (1),
\[C=1-A\]
Substituting this into equation (2),
\[ \begin{aligned} 3A+B+2(1-A)&=1\\ 3A+B+2-2A&=1\\ A+B&=-1 \end{aligned} \]
Therefore,
\[B=-1-A \tag{4}\]
Substituting \(C=1-A\) into equation (3),
\[ \begin{aligned} 2A+2B+(1-A)&=1\\ A+2B+1&=1\\ A+2B&=0 \end{aligned} \]
Using equation (4),
\[ \begin{aligned} A+2(-1-A)&=0\\ A-2-2A&=0\\ -A&=2 \end{aligned} \]
Hence,
\[A=-2\]
From equation (4),
\[B=-1-(-2)=1\]
Also,
\[C=1-(-2)=3\]
Thus,
\[\boxed{A=-2,\quad B=1,\quad C=3}\]
Write the partial fraction decomposition
Substituting the values of \(A\), \(B\) and \(C\),
\[\boxed{\frac{x^2+x+1}{(x+1)^2(x+2)}=-\frac{2}{x+1}+\frac{1}{(x+1)^2}+\frac{3}{x+2}}\]
Integrate term by term
Substituting the decomposition into the original integral,
\[\begin{aligned}I&=\int\left[-\frac{2}{x+1}+\frac{1}{(x+1)^2}+\frac{3}{x+2}\right]dx\\ &=-2\int\frac{dx}{x+1}+\int\frac{dx}{(x+1)^2}+3\int\frac{dx}{x+2}\end{aligned}\]
Using the standard integration formulas,
\[\begin{aligned}I&=-2\log|x+1|-\frac{1}{x+1}+3\log|x+2|+C\end{aligned}\]

💡 Answer

Final Answer

›

\[ \boxed{ \begin{aligned} \int\frac{x^2+x+1}{(x+1)^2(x+2)}\,dx ={}&-2\log|x+1|\\ &-\frac{1}{x+1}\\ &+3\log|x+2|+C \end{aligned} } \]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem tests partial fraction decomposition involving repeated linear factors and the integration of rational functions.
JEE Main: Identifying the correct decomposition and determining its coefficients are essential skills for evaluating rational integrals.
JEE Advanced: The problem reinforces algebraic manipulation and the treatment of repeated factors in partial fraction decomposition.
CUET and other entrance examinations: It tests the application of logarithmic integration and the standard power rule for negative exponents.

📜 Note

Important Notes

›

The original integrand is undefined at

\[ x=-1,\qquad x=-2 \]

Therefore, the antiderivative is considered separately on the intervals

\[ (-\infty,-2),\quad(-2,-1),\quad(-1,\infty) \]

The integration constant may be different on each connected interval of the domain.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For repeated linear factors, include every power of the repeated factor in the partial fraction decomposition.
The correct decomposition here is \(-\frac{2}{x+1}+\frac{1}{(x+1)^2}+\frac{3}{x+2}\).
Remember that \(\int\frac{dx}{(x+a)^2}=-\frac{1}{x+a}+C\).
Logarithmic integration requires absolute values when the sign of the linear expression is unrestricted.
Verification by differentiation is an effective way to detect errors in partial fraction coefficients and integration signs.

📘 Concept & Theory

Theory and Concept

›

This problem involves two important concepts.

Concept 1: Inverse trigonometric identity

For \(-1

\[ \tan\left(\frac{\theta}{2}\right) =\sqrt{\frac{1-\cos\theta}{1+\cos\theta}} \]

gives

\[ \tan^{-1}\sqrt{\frac{1-x}{1+x}} =\frac12\cos^{-1}x \]

Concept 2: Integration by parts

Recall the formula

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

We use this formula with the inverse cosine function as the first function and \(dx\) as the second function.

🗺️ Solution Roadmap

Step-by-step Plan

›

Recognise the half-angle identity hidden in the given expression.
Convert the inverse tangent expression into an inverse cosine function.
Apply integration by parts.
Evaluate the remaining standard integral.
Combine the terms and verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 15 steps

Apply the inverse trigonometric identity
We know that
\[\cos 2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}\]
Rearranging,
\[\tan^2\theta=\frac{1-\cos2\theta}{1+\cos2\theta}\]
Taking the positive square root,
\[\tan\theta=\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}\]
Comparing this with the given expression, we obtain
\[\theta=\frac12\cos^{-1}x\]
Therefore,
\[\tan^{-1}\sqrt{\frac{1-x}{1+x}}=\frac12\cos^{-1}x\]
Hence, the integral becomes
\[I=\frac12\int\cos^{-1}x\,dx\]
Apply integration by parts
Let
\[u=\cos^{-1}x,\quad dv=dx\]
Differentiating and integrating respectively,
\[du=-\frac{dx}{\sqrt{1-x^2}},\quad v=x\]
Using
\[\int u\,dv=uv-\int v\,du,\]
we get
\[\begin{aligned}I&=\frac12\left[x\cos^{-1}x-\int x\left(-\frac{1}{\sqrt{1-x^2}}\right)dx\right]\\ &=\frac12\left[x\cos^{-1}x+\int\frac{x}{\sqrt{1-x^2}}\,dx\right]\end{aligned}\]
Evaluate the remaining integral
Let
\[t=1-x^2\]
Differentiating,
\[dt=-2x\,dx\]
Therefore,
\[x\,dx=-\frac12dt\]
Substituting,
\[ \begin{aligned} \int\frac{x}{\sqrt{1-x^2}}\,dx &=-\frac12\int t^{-1/2}\,dt\\ &=-\frac12\cdot2t^{1/2}+C\\ &=-\sqrt{t}+C\\ &=-\sqrt{1-x^2}+C \end{aligned} \]
Substitute the result back
Thus,
\[\begin{aligned}I&=\frac12\left[x\cos^{-1}x-\sqrt{1-x^2}\right]+C\end{aligned}\]

💡 Answer

Final Answer

›

\[\boxed{\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}\,dx=\frac12\left[x\cos^{-1}x-\sqrt{1-x^2}\right]+C}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: This problem tests inverse trigonometric identities and integration by parts.
JEE Main: Recognising the half-angle identity simplifies a complicated inverse tangent expression into an inverse cosine function.
JEE Advanced: The problem demonstrates how an appropriate trigonometric transformation can reduce a non-standard integral to a standard form.
CUET and other entrance examinations: It reinforces inverse trigonometric identities, differentiation of inverse functions and integration by parts.

📜 Note

Important Notes

›

For the original expression to be real-valued, the radicand must be non-negative and the denominator must be positive. Thus, its real domain is \(-1<x\leq1\).
The identity used above is valid for \(-1<x\leq1\), with the inverse cosine taken in its principal range \([0,\pi]\).
The integration formula is derived on \(-1<x<1\); the endpoint \(x=1\) can be handled by continuity.
The integration constant is understood on the connected interval of the domain.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Recognise the half-angle identity \(\tan(\theta/2)=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\) when the principal angle lies in the appropriate range.
Convert complicated inverse trigonometric expressions into simpler forms before integrating.
Use integration by parts for integrals involving inverse trigonometric functions.
The integral \(\int\frac{x}{\sqrt{1-x^2}}\,dx=-\sqrt{1-x^2}+C\) is a useful standard result.
Differentiate the final expression to verify the antiderivative.

📘 Concept & Theory

Theory and Concept

›

We use the following concepts:

Logarithmic properties: Combining logarithms to simplify the integrand.
Integration by parts: Differentiating the logarithmic factor while integrating the remaining algebraic factor.
Hyperbolic substitution: Using \(x=\sinh t\) to simplify expressions involving \(\sqrt{x^2+1}\).
Dilogarithm function: The special function \(\operatorname{Li}_2(z)\), defined by

\[ \operatorname{Li}_2(z) =-\int_0^z\frac{\log(1-u)}{u}\,du \]

In particular,

\[ \frac{d}{dz}\operatorname{Li}_2(z) =-\frac{\log(1-z)}{z} \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Combine the logarithms into a single logarithmic expression.
Apply integration by parts using \(\sqrt{x^2+1}\) as the algebraic factor.
Transform the remaining logarithmic integral using \(x=\sinh t\).
Introduce the substitution \(q=\tanh(t/2)\).
Evaluate the resulting integral using the dilogarithm function.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 35 steps

Simplify the logarithmic expression
Using the logarithmic identity,
\[\log a-\log b=\log\left(\frac ab\right),\]
we obtain
\[\log(x^2+1)-2\log x=\log\left(\frac{x^2+1}{x^2}\right)\]
Therefore,
\[I=\int\frac{\sqrt{x^2+1}}{x}\log\left(\frac{x^2+1}{x^2}\right)dx\]
For convenience, let
\[L=\log\left(\frac{x^2+1}{x^2}\right)\]
Thus,
\[I=\int\frac{\sqrt{x^2+1}}{x}L\,dx\]
Apply integration by parts
Choose
\[u=L,\quad dv=\frac{\sqrt{x^2+1}}{x}\,dx\]
Differentiating \(u\),
\[ \begin{aligned} du &=\left(\frac{2x}{x^2+1}-\frac2x\right)dx\\ &=-\frac{2}{x(x^2+1)}\,dx \end{aligned} \]
Now, integrating \(dv\),
\[v=\int\frac{\sqrt{x^2+1}}{x}\,dx\]
We know that
\[v=\sqrt{x^2+1}+\log\left(\frac{x}{1+\sqrt{x^2+1}}\right)\]
Using the integration-by-parts formula,
\[I=uv-\int v\,du\]
Hence,
\[\begin{aligned}I={}&L\left[\sqrt{x^2+1}+\log\left(\frac{x}{1+\sqrt{x^2+1}}\right)\right]\\ &+2\int\frac{1}{x(x^2+1)}\left[\sqrt{x^2+1}+\log\left(\frac{x}{1+\sqrt{x^2+1}}\right)\right]dx\end{aligned}\]
After simplifying and rearranging the terms, an equivalent and more convenient decomposition is
\[I=\sqrt{x^2+1}\,L+\int\frac{L}{x\sqrt{x^2+1}}\,dx+2\int\frac{dx}{x\sqrt{x^2+1}}\]
Evaluate the elementary integral
We have
\[K=\int\frac{dx}{x\sqrt{x^2+1}}\]
Using the standard integral,
\[K=\log\left(\frac{x}{1+\sqrt{x^2+1}}\right)\]
Introduce the substitution
\[q=\frac{x}{1+\sqrt{x^2+1}}\]
Therefore,
\[K=\log q\]
Evaluate the remaining logarithmic integral
Let
\[J=\int\frac{L}{x\sqrt{x^2+1}}\,dx\]
Put
\[x=\sinh t\]
Then,
\[\sqrt{x^2+1}=\cosh t,\quad dx=\cosh t\,dt\]
Consequently,
\[\frac{dx}{x\sqrt{x^2+1}}=\frac{dt}{\sinh t}\]
Also,
\[\begin{aligned}L&=\log\left(\frac{\cosh^2t}{\sinh^2t}\right)\\ &=2\log(\coth t)\end{aligned}\]
Thus,
\[J=2\int\log(\coth t)\,\operatorname{csch}t\,dt\]
Now, introduce
\[q=\tanh\frac t2\]
Using the hyperbolic identities,
\[\operatorname{csch}t\,dt=\frac{dq}{q}\]
and
\[\coth t=\frac{1+q^2}{2q}\]
Therefore,
\[J=2\int\log\left(\frac{1+q^2}{2q}\right)\frac{dq}{q}\]
Expanding the logarithm,
\[J=2\int\left[\log(1+q^2)-\log2-\log q\right]\frac{dq}{q}\]
Separating the integrals,
\[ \begin{aligned} J={}&2\int\frac{\log(1+q^2)}q\,dq\\ &-2\log2\int\frac{dq}{q} -2\int\frac{\log q}{q}\,dq \end{aligned} \]
From the definition of the dilogarithm,
\[\frac{d}{dq}\operatorname{Li}_2(-q^2)=-\frac{2\log(1+q^2)}q\]
Hence,
\[2\int\frac{\log(1+q^2)}q\,dq=-\operatorname{Li}_2(-q^2)\]
Also,
\[\int\frac{dq}{q}=\log q\]
and
\[\int\frac{\log q}{q}\,dq=\frac12(\log q)^2\]
Substituting these results,
\[\boxed{J=-\operatorname{Li}_2(-q^2)-2\log2\log q-(\log q)^2+C}\]
Combine all the results
Recall that
\[I=\sqrt{x^2+1}\,L+J+2K\]
Substituting the expressions for \(J\) and \(K\),
\[\begin{aligned}I={}&\sqrt{x^2+1}\,L-\operatorname{Li}_2(-q^2)\\ &-2\log2\log q-(\log q)^2+2\log q+C\end{aligned}\]
Finally, substituting
\[L=\log\left(\frac{x^2+1}{x^2}\right)\]
and
\[q=\frac{x}{1+\sqrt{x^2+1}},\]
\[\boxed{\begin{aligned}I={}&\sqrt{x^2+1}\log\left(\frac{x^2+1}{x^2}\right)\\ &-\operatorname{Li}_2\left(-\frac{x^2}{(1+\sqrt{x^2+1})^2}\right)\\ &-2\log2\log\left(\frac{x}{1+\sqrt{x^2+1}}\right)\\ &-\left[\log\left(\frac{x}{1+\sqrt{x^2+1}}\right)\right]^2\\ &+2\log\left(\frac{x}{1+\sqrt{x^2+1}}\right)+C\end{aligned}}\]

💡 Answer

Final Answer

›

\[\boxed{\begin{aligned}I={}&\sqrt{x^2+1}\log\left(\frac{x^2+1}{x^2}\right)\\ &-\operatorname{Li}_2\left(-\frac{x^2}{(1+\sqrt{x^2+1})^2}\right)\\ &-2\log2\log\left(\frac{x}{1+\sqrt{x^2+1}}\right)\\ &-\left[\log\left(\frac{x}{1+\sqrt{x^2+1}}\right)\right]^2\\ &+2\log\left(\frac{x}{1+\sqrt{x^2+1}}\right)+C\end{aligned}}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: The problem reinforces logarithmic simplification and integration by parts.
JEE Main and Advanced: It illustrates how a complicated integral can lead to a special function rather than an elementary antiderivative.
Higher Mathematics: The dilogarithm function appears in advanced integration, mathematical analysis, number theory and mathematical physics.

📜 Note

Important Notes

›

The original expression contains \(\log x\), so its real domain is \(x>0\).
The dilogarithm \(\operatorname{Li}_2(z)\) is also called the Spence function.
This integral is not an ordinary elementary-function integration problem. Its antiderivative naturally involves the dilogarithm.
The integration constant is arbitrary on the interval \(x>0\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Combine logarithmic terms before choosing a substitution or integration technique.
Integration by parts can reduce an integral to a more manageable combination of terms.
The substitution \(x=\sinh t\) is useful for integrals involving \(\sqrt{x^2+1}\).
The substitution \(q=\tanh(t/2)\) converts the remaining hyperbolic integral into a logarithmic integral.
Not every integral has an elementary antiderivative; some require special functions such as the dilogarithm.

📘 Concept & Theory

Theory and Concept

›

We use the tangent half-angle substitution:

\[ t=\tan\frac{x}{2} \]

Recall the identities:

\[ \sin x=\frac{2t}{1+t^2}, \qquad \cos x=\frac{1-t^2}{1+t^2} \]

\[ dx=\frac{2}{1+t^2}\,dt \]

This substitution converts the trigonometric quotient into a rational function of \(t\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Put \(t=\tan\frac{x}{2}\).
Transform the trigonometric quotient and the differential.
Change the limits of integration.
Evaluate the resulting definite integral numerically.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Apply the half-angle substitution
Let
\[t=\tan\frac{x}{2}\]
Then,
\[\sin x=\frac{2t}{1+t^2}\]

\[\cos x=\frac{1-t^2}{1+t^2}\]
Therefore,
\[ \begin{aligned} \frac{1-\sin x}{1-\cos x} &= \frac{1-\dfrac{2t}{1+t^2}} {1-\dfrac{1-t^2}{1+t^2}}\\[6pt] &= \frac{\dfrac{(t-1)^2}{1+t^2}} {\dfrac{2t^2}{1+t^2}}\\[6pt] &=\frac{(t-1)^2}{2t^2} \end{aligned} \]
Also,
\[dx=\frac{2}{1+t^2}\,dt\]
Change the limits
When \(x=\frac{\pi}{2}\),
\[t=\tan\frac{\pi}{4}=1\]
When \(x=\pi\),
\[t=\tan\frac{\pi}{2}\longrightarrow\infty\]
Thus, the given integral becomes
\[ \begin{aligned} I &=\int_1^\infty e^{2\tan^{-1}t} \frac{(t-1)^2}{2t^2} \frac{2}{1+t^2}\,dt\\[6pt] &=\int_1^\infty \frac{e^{2\tan^{-1}t}(t-1)^2} {t^2(1+t^2)}\,dt \end{aligned} \]
Evaluate the transformed integral
The resulting integral does not reduce to a standard elementary antiderivative. It can be evaluated using numerical quadrature.
\[I=\int_1^\infty\frac{e^{2\tan^{-1}t}(t-1)^2}{t^2(1+t^2)}\,dt\]
Numerical evaluation gives
\[I\approx 4.810477380965\]
\[\boxed{I\approx 4.81048}\]

💡 Answer

Final Answer

›

\[\boxed{I\approx 4.81048}\]

🎯 Exam Significance

Exam Significance

›

Trigonometric substitution: The tangent half-angle substitution is useful for converting rational expressions in sine and cosine into rational functions.
Definite integrals: Changing the limits correctly is essential when substituting variables.
Numerical integration: Numerical quadrature is useful when an integral does not have a convenient elementary antiderivative.

📜 Note

Important Notes

›

The original integrand is continuous on the entire interval \(\left[\frac{\pi}{2},\pi\right]\), since \(1-\cos x\ne0\) there.
The transformed upper limit is an improper limit because \(\tan\frac{x}{2}\to\infty\) as \(x\to\pi^{-}\). The integral nevertheless converges.
The numerical value is for the expression exactly as supplied. If this question is from an NCERT exercise or answer key that expects an elementary exact value, it is worth checking whether the numerator or denominator has been transcribed correctly.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Use \(t=\tan\frac{x}{2}\) to simplify rational expressions involving \(\sin x\) and \(\cos x\).
Transform both the integrand and the limits.
Do not assume that every definite integral has an elementary exact value.
Check the original question carefully if an elementary answer is expected but the transformed integral does not simplify.

📘 Concept & Theory

Theory and Concept

›

We use the substitution

\[ t=\tan x \]

Recall that

\[ \tan x=\frac{\sin x}{\cos x}, \qquad dt=\sec^2x\,dx \]

Also,

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

and

\[ \cos^4x+\sin^4x =\cos^4x(1+\tan^4x) \]

This substitution simplifies the trigonometric expression into a rational function.

🗺️ Solution Roadmap

Step-by-step Plan

›

Put \(t=\tan x\).
Express the numerator, denominator and differential in terms of \(t\).
Change the limits of integration.
Use the substitution \(u=t^2\) to evaluate the resulting integral.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 13 steps

Apply the substitution
Let
\[t=\tan x\]
Then,
\[dt=\sec^2x\,dx=\frac{dx}{\cos^2x}\]
Therefore,
\[dx=\frac{dt}{1+t^2}\]
Also,
\[\sin x\cos x=\frac{t}{1+t^2}\]
and
\[ \begin{aligned} \cos^4x+\sin^4x &=\cos^4x(1+\tan^4x)\\ &=\frac{1+t^4}{(1+t^2)^2} \end{aligned} \]
Change the limits
When \(x=0\),
\[t=\tan0=0\]
When \(x=\frac{\pi}{4}\),
\[t=\tan\frac{\pi}{4}=1\]
Substituting these expressions into the integral,
\[ \begin{aligned} I &=\int_0^1 \frac{\dfrac{t}{1+t^2}} {\dfrac{1+t^4}{(1+t^2)^2}} \cdot\frac{dt}{1+t^2}\\[8pt] &=\int_0^1\frac{t}{1+t^4}\,dt \end{aligned} \]
Evaluate the simplified integral.
Put
\[u=t^2\]
Then,
\[du=2t\,dt\]
Therefore,
\[t\,dt=\frac12 du\]
Changing the limits:
\[ \begin{aligned} t=0&\implies u=0\\ t=1&\implies u=1 \end{aligned} \]
Thus,
\[ \begin{aligned} I &=\frac12\int_0^1\frac{du}{1+u^2}\\[6pt] &=\frac12\left[\tan^{-1}u\right]_0^1\\[6pt] &=\frac12\left(\frac{\pi}{4}-0\right)\\[6pt] &=\frac{\pi}{8} \end{aligned} \]
\[\boxed{\int_0^{\pi/4}\frac{\sin x\cos x}{\cos^4x+\sin^4x}\,dx=\frac{\pi}{8}}\]

💡 Answer

Final Answer

›

\[\boxed{\int_0^{\pi/4}\frac{\sin x\cos x}{\cos^4x+\sin^4x}\,dx=\frac{\pi}{8}}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: Tests substitution and evaluation of definite integrals.
JEE Main: Demonstrates the usefulness of \(t=\tan x\) in simplifying trigonometric rational expressions.
CUET and other entrance examinations: Reinforces the conversion of trigonometric integrals into standard inverse-tangent integrals.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

The substitution \(t=\tan x\) simplifies expressions involving powers of sine and cosine.
Always transform the differential and both limits when evaluating a definite integral by substitution.
The identity \(\cos^4x+\sin^4x=\cos^4x(1+\tan^4x)\) is useful in this problem.
The standard integral \(\int\frac{du}{1+u^2}=\tan^{-1}u+C\) completes the evaluation.

📘 Concept & Theory

Theory and Concept

›

When an integral involves a rational expression in \(\sin x\) and \(\cos x\), the substitution

\[ t=\tan x \]

is often useful because it converts the trigonometric functions into rational functions of \(t\). We use the identities

\[ \cos^2x=\frac{1}{1+t^2},\qquad \sin^2x=\frac{t^2}{1+t^2} \]

\[ dx=\frac{dt}{1+t^2} \]

🗺️ Solution Roadmap

Step-by-step Plan

›

We will substitute \(t=\tan x\), transform the limits, simplify the integrand into a rational function, and evaluate the resulting improper integral using partial fractions.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Apply the substitution
\[t=\tan x\]
Differentiating both sides,
\[\sec^2x\,dx=dt\]

\[dx=\frac{dt}{1+\tan^2x}=\frac{dt}{1+t^2}\]
Also,
\[\cos^2x=\frac{1}{1+t^2},\qquad\sin^2x=\frac{t^2}{1+t^2}\]
Transform the limits
When \(x=0\), \(t=\tan 0=0\)
When \(x=\dfrac{\pi}{2}\), \(t=\tan\dfrac{\pi}{2}\to\infty\).
Therefore,
\[I=\int_0^\infty\frac{\dfrac{1}{1+t^2}}{\dfrac{1}{1+t^2}+\dfrac{4t^2}{1+t^2}}\cdot\frac{dt}{1+t^2}\]
Simplify the integrand
\[I=\int_0^\infty\frac{1}{1+4t^2}\cdot\frac{dt}{1+t^2}\]

\[I=\int_0^\infty\frac{dt}{(1+t^2)(1+4t^2)}\]
Resolve into partial fractions
Let
\[\frac{1}{(1+t^2)(1+4t^2)}=\frac{A}{1+t^2}+\frac{B}{1+4t^2}\]
Taking the LCM and comparing coefficients,
\[1=A(1+4t^2)+B(1+t^2)\]

\[1=(A+B)+(4A+B)t^2\]
Comparing the constant and \(t^2\) coefficients,
\[A+B=1,\quad 4A+B=0\]
Subtracting the first equation from the second,
\[3A=-1\implies A=-\frac13\]

\[B=1-A=1+\frac13=\frac43\]
Thus,
\[I=\int_0^\infty\left[-\frac{1}{3(1+t^2)}+\frac{4}{3(1+4t^2)}\right]dt\]
Integrate
Using the standard result
\[\int\frac{dt}{1+a^2t^2}=\frac{1}{a}\tan^{-1}(at)+C\]
we obtain
\[ I= -\frac13\left[\tan^{-1}t\right]_0^\infty +\frac43\left[\frac12\tan^{-1}(2t)\right]_0^\infty \]

\[ I= -\frac13\left(\frac{\pi}{2}-0\right) +\frac23\left(\frac{\pi}{2}-0\right) \]

\[ I=-\frac{\pi}{6}+\frac{\pi}{3} \]

\[ I=\frac{\pi}{6} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int_0^{\frac{\pi}{2}}\frac{\cos^2x}{\cos^2x+4\sin^2x}\,dx=\frac{\pi}{6}}\]

🎯 Exam Significance

Exam Significance

›

CBSE Boards: Tests the substitution \(t=\tan x\), transformation of definite-integral limits, and partial-fraction decomposition.
JEE Main: Reinforces the conversion of trigonometric integrals into rational integrals.
CUET: Tests standard inverse-tangent integration formulas and evaluation at infinite limits.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

For rational expressions involving \(\sin x\) and \(\cos x\), consider \(t=\tan x\).
Always transform the limits when changing the variable in a definite integral.
When the upper limit becomes infinity, evaluate the integral as an improper integral using the limiting value.
Remember that \(\displaystyle\lim_{t\to\infty}\tan^{-1}(at)=\frac{\pi}{2}\) for \(a>0\).

📘 Concept & Theory

Theory and Concept

›

We use the modulus identity:

\[ \sqrt{a^2}=|a| \]

Thus,

\[ \sqrt{\sin^2x}=|\sin x| \]

Since \(\sin x>0\) throughout the given interval,

\[ |\sin x|=\sin x \]

We also use the standard integrals:

\[ \int 1\,dx=x+C \]

\[ \int\cot x\,dx=\log|\sin x|+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Simplify the square root using the sign of \(\sin x\) in the given interval.
Separate the integrand into two standard terms.
Integrate and apply the given limits.
Simplify the logarithmic expression.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Given —
\[I=\int_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin^2x}}\,dx\]
Since
\[\frac{\pi}{6}\leq x\leq\frac{\pi}{3}\]
we have \(\sin x>0\). Hence,
\[\sqrt{\sin^2x}=\sin x\]
Therefore,
\[ \begin{aligned} I &=\int_{\pi/6}^{\pi/3} \frac{\sin x+\cos x}{\sin x}\,dx\\[6pt] &=\int_{\pi/6}^{\pi/3}(1+\cot x)\,dx \end{aligned} \]
Integrate the expression
Using the standard integrals,
\[ \begin{aligned} I &=\int_{\pi/6}^{\pi/3}1\,dx +\int_{\pi/6}^{\pi/3}\cot x\,dx\\[6pt] &=\left[x+\log|\sin x|\right]_{\pi/6}^{\pi/3} \end{aligned} \]
Apply the limits
\[ \begin{aligned} I ={}&\left[ \frac{\pi}{3}+\log\left(\sin\frac{\pi}{3}\right) \right]\\ &-\left[ \frac{\pi}{6}+\log\left(\sin\frac{\pi}{6}\right) \right] \end{aligned} \]
Using the standard trigonometric values,
\[ \sin\frac{\pi}{3}=\frac{\sqrt3}{2}, \quad \sin\frac{\pi}{6}=\frac12 \]
We get
\[ \begin{aligned} I &=\frac{\pi}{3}-\frac{\pi}{6} +\log\frac{\sqrt3}{2} -\log\frac12\\[6pt] &=\frac{\pi}{6} +\log\left(\frac{\sqrt3/2}{1/2}\right)\\[6pt] &=\frac{\pi}{6}+\log\sqrt3 \end{aligned} \]
Since \(\log\sqrt3=\frac12\log3\),
\[I=\frac{\pi}{6}+\frac12\log3\]
\[\boxed{I=\frac{\pi}{6}+\frac12\log3}\]

💡 Answer

Final Answer

›

\[\boxed{I=\frac{\pi}{6}+\frac12\log3}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: Tests the evaluation of definite integrals using standard antiderivatives and trigonometric identities.
JEE Main: Highlights the importance of correctly interpreting square roots of squared trigonometric functions.
CUET and other entrance examinations: Reinforces the use of logarithmic integration and exact trigonometric values.

📜 Note

Important Notes

›

Remember that

\[ \sqrt{\sin^2x}=|\sin x| \]

It is not always equal to \(\sin x\). The simplification to \(\sin x\) is valid here specifically because the entire interval lies in the first quadrant.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Always use \(\sqrt{a^2}=|a|\), not simply \(a\).
Check the sign of the trigonometric function over the complete interval before simplifying.
Recognise that \(\frac{\sin x+\cos x}{\sin x}=1+\cot x\).
Apply the limits only after finding the complete antiderivative.
The exact value of the integral is \(\frac{\pi}{6}+\frac12\log3\).

📘 Concept & Theory

Theory and Concept

›

We use the identity

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

Multiplying the numerator and denominator by the conjugate of the denominator rationalises the expression.

We also use the standard integral:

\[ \int\sqrt{x}\,dx=\frac{2}{3}x^{3/2}+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Multiply the numerator and denominator by the conjugate expression.
Simplify the resulting integrand.
Integrate the simplified expression.
Substitute the upper and lower limits.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 4 steps

Given —
\[I=\int_0^1\frac{dx}{\sqrt{1+x}-\sqrt{x}}\]
Rationalise the denominator by Multiplying the numerator and denominator by the conjugate \(\sqrt{1+x}+\sqrt{x}\),
\[ \begin{aligned} I &=\int_0^1 \frac{\sqrt{1+x}+\sqrt{x}} {(\sqrt{1+x}-\sqrt{x})(\sqrt{1+x}+\sqrt{x})}\,dx\\[6pt] &=\int_0^1 \frac{\sqrt{1+x}+\sqrt{x}} {(1+x)-x}\,dx\\[6pt] &=\int_0^1\left[\sqrt{1+x}+\sqrt{x}\right]dx \end{aligned} \]
Integrate the expression
Separating the integrals,
\[I=\int_0^1\sqrt{1+x}\,dx+\int_0^1\sqrt{x}\,dx\]
Using the power rule of integration,
\[ \begin{aligned} I &=\left[\frac{2}{3}(1+x)^{3/2}\right]_0^1 +\left[\frac{2}{3}x^{3/2}\right]_0^1\\[6pt] &=\frac{2}{3}\left[(1+x)^{3/2}+x^{3/2}\right]_0^1 \end{aligned} \]
Apply the limits
\[ \begin{aligned} I &=\frac{2}{3} \left[ (2)^{3/2}+1^{3/2} -\left(1^{3/2}+0^{3/2}\right) \right]\\[6pt] &=\frac{2}{3}\left[2\sqrt2+1-1\right]\\[6pt] &=\frac{4\sqrt2}{3} \end{aligned} \]

💡 Answer

Final Answer

›

\[ \boxed{ \int_0^1\frac{dx}{\sqrt{1+x}-\sqrt{x}} =\frac{4\sqrt2}{3} } \]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: Tests rationalisation and the application of standard integration formulas.
JEE Main: Demonstrates how multiplying by a conjugate can convert a complicated radical expression into a simple sum.
CUET and other entrance examinations: Reinforces definite integration, fractional powers and the correct application of limits.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

When the denominator contains a difference of square roots, consider multiplying by its conjugate.
The identity \((a-b)(a+b)=a^2-b^2\) eliminates the radical denominator.
Use the power rule to integrate the resulting fractional powers.
Substitute the limits carefully to obtain the exact value.

📘 Concept & Theory

Theory and Concept

›

We use the substitution method for definite integration. The key observation is

\[ \frac{d}{dx}(\sin x-\cos x)=\cos x+\sin x \]

We also use the identity

\[ (\sin x-\cos x)^2=1-\sin 2x \]

These two relationships allow us to express the entire integrand in terms of a single variable.

🗺️ Solution Roadmap

Step-by-step Plan

›

Choose \(t=\sin x-\cos x\), using the numerator as its derivative.
Express \(\sin 2x\) in terms of \(t\).
Transform the limits of integration.
Evaluate the resulting rational integral.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Apply the substitution
Let
\[t=\sin x-\cos x\]
Differentiating,
\[dt=(\cos x+\sin x)\,dx\]
Therefore,
\[(\sin x+\cos x)\,dx=dt\]
Express the denominator in terms of \(t\)
Squaring the substitution,
\[ \begin{aligned} t^2 &=(\sin x-\cos x)^2\\ &=\sin^2x+\cos^2x-2\sin x\cos x\\ &=1-\sin 2x \end{aligned} \]
Hence,
\[\sin 2x=1-t^2\]
Substituting into the denominator,
\[ \begin{aligned} 9+16\sin 2x &=9+16(1-t^2)\\ &=25-16t^2 \end{aligned} \]
Thus, the integral becomes
\[I=\int\frac{dt}{25-16t^2}\]
Change the limits of integration
When \(x=0\),
\[t=\sin0-\cos0=-1\]
When \(x=\frac{\pi}{4}\),
\[t=\sin\frac{\pi}{4}-\cos\frac{\pi}{4}=0\]
Therefore,
\[I=\int_{-1}^{0}\frac{dt}{25-16t^2}\]
Evaluate the integral
Using the standard formula
\[ \int\frac{du}{a^2-b^2u^2} =\frac{1}{2ab} \log\left|\frac{a+bu}{a-bu}\right|+C \]
we obtain
\[ \begin{aligned} I &=\left[ \frac{1}{40} \log\left|\frac{5+4t}{5-4t}\right| \right]_{-1}^{0}\\[6pt] &=\frac{1}{40} \left[ \log 1-\log\left(\frac{1}{9}\right) \right]\\[6pt] &=\frac{1}{40}\log9\\[6pt] &=\frac{1}{20}\log3 \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\int_0^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x}\,dx=\frac{1}{20}\log3}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: Tests substitution in definite integrals and the use of standard logarithmic integration formulas.
JEE Main: Demonstrates how recognising the derivative of a suitable trigonometric expression simplifies an integral.
CUET and other entrance examinations: Reinforces trigonometric identities, change of limits and rational integration.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Look for a substitution whose derivative matches the numerator.
The identity \((\sin x-\cos x)^2=1-\sin2x\) is central to this problem.
Always transform the limits when changing variables in a definite integral.
The standard integral involving \(a^2-b^2t^2\) produces a logarithmic result.

📘 Concept & Theory

Theory and Concept

›

We use the integration-by-parts formula:

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

We also use the identity:

\[ \sin 2x=2\sin x\cos x \]

and the derivative:

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

🗺️ Solution Roadmap

Step-by-step Plan

›

Apply integration by parts by choosing the inverse tangent expression as the first function.
Evaluate the resulting integral using the substitution \(t=\sin x\).
Use a standard logarithmic integral to obtain the exact result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Apply integration by parts
We have
\[I=\int_0^{\pi/2}\sin 2x\,\tan^{-1}(\sin x)\,dx\]
Let
\[u=\tan^{-1}(\sin x)\]
and
\[dv=\sin 2x\,dx\]
Then,
\[du=\frac{\cos x}{1+\sin^2x}\,dx\]
and
\[v=-\frac12\cos 2x\]
Using integration by parts,
\[I=\left[-\frac12\cos 2x\,\tan^{-1}(\sin x)\right]_0^{\pi/2}+\frac12\int_0^{\pi/2}\frac{\cos 2x\cos x}{1+\sin^2x}\,dx\]
Evaluate the boundary term
At \(x=\frac{\pi}{2}\),
\[-\frac12\cos\pi\,\tan^{-1}(1)=\frac{\pi}{8}\]
At \(x=0\),
\[-\frac12\cos0\,\tan^{-1}(0)=0\]
Therefore,
\[I=\frac{\pi}{8}+\frac12\int_0^{\pi/2}\frac{\cos 2x\cos x}{1+\sin^2x}\,dx\]
Use substitution
Let
\[t=\sin x\]
Then,
\[dt=\cos x\,dx\]
Also,
\[\cos 2x=1-2\sin^2x=1-2t^2\]
Changing the limits:
\[ \begin{aligned} x=0&\implies t=0\\ x=\frac{\pi}{2}&\implies t=1 \end{aligned} \]
Thus,
\[I=\frac{\pi}{8}+\frac12\int_0^1\frac{1-2t^2}{1+t^2}\,dt\]
Simplify and integrate
We can write
\[\begin{aligned}\frac{1-2t^2}{1+t^2}&=\frac{3}{1+t^2}-2\end{aligned}\]
Therefore,
\[ \begin{aligned} I &=\frac{\pi}{8} +\frac12\int_0^1 \left(\frac{3}{1+t^2}-2\right)dt\\[6pt] &=\frac{\pi}{8} +\frac12\left[ 3\tan^{-1}t-2t \right]_0^1\\[6pt] &=\frac{\pi}{8} +\frac12\left(\frac{3\pi}{4}-2\right)\\[6pt] &=\frac{\pi}{8}+\frac{3\pi}{8}-1\\[6pt] &=\frac{\pi}{2}-1 \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\int_0^{\pi/2}\sin 2x\,\tan^{-1}(\sin x)\,dx=\frac{\pi}{2}-1}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12 Mathematics: Tests integration by parts and substitution in definite integrals.
JEE Main: Demonstrates how to combine inverse trigonometric differentiation with trigonometric identities.
CUET and other entrance examinations: Reinforces the evaluation of definite integrals through suitable substitutions and standard inverse-tangent integrals.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Integration by parts is effective when an inverse trigonometric function is multiplied by a trigonometric function.
The derivative of \(\tan^{-1}(\sin x)\) is \(\frac{\cos x}{1+\sin^2x}\).
The substitution \(t=\sin x\) simplifies integrals containing \(\cos x\,dx\).
Use \(\cos 2x=1-2\sin^2x\) to express the remaining integrand in terms of \(t\).
The exact value of the given definite integral is \(\frac{\pi}{2}-1\).

📘 Concept & Theory

Theory and Concept

›

When evaluating definite integrals involving absolute-value functions, we must first identify the points at which the expressions inside the modulus change sign.

Recall that

\[ |x-a|= \begin{cases} a-x, & x

Therefore, the interval of integration must be divided at the zeros of the expressions inside the modulus. On each resulting subinterval, the absolute-value expressions can be replaced by their appropriate algebraic forms.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the critical points of the absolute-value expressions.
Divide the interval \([1,4]\) at those points.
Remove the modulus signs separately on each subinterval.
Evaluate the resulting polynomial integrals and add their values.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Given —
\[I=\int_1^4\left[|x-1|+|x-2|+|x-3|\right]dx\]
Identify the critical points
The expressions inside the modulus signs become zero at
\[x-1=0\Rightarrow x=1\]

\[x-2=0\Rightarrow x=2\]

\[x-3=0\Rightarrow x=3\]
Thus, divide the interval \([1,4]\) into three subintervals:
\[[1,2],\quad [2,3],\quad [3,4]\]
Remove the modulus signs on each subinterval
For \(1\leq x\leq 2\):
Here, \(x-1\geq 0\), while \(x-2\leq 0\) and \(x-3\leq 0\). Hence,
\[|x-1|=x-1\]

\[|x-2|=2-x\]

\[|x-3|=3-x\]
Adding these expressions,
\[ \begin{aligned} |x-1|+|x-2|+|x-3| &=(x-1)+(2-x)+(3-x)\\ &=4-x. \end{aligned} \]
For \(2\leq x\leq 3\):
Here, \(x-1\geq 0\), \(x-2\geq 0\), and \(x-3\leq 0\). Therefore,
\[ \begin{aligned} |x-1|+|x-2|+|x-3| &=(x-1)+(x-2)+(3-x)\\ &=x. \end{aligned} \]
For \(3\leq x\leq 4\):
All three expressions inside the modulus signs are non-negative. Thus,
\[ \begin{aligned} |x-1|+|x-2|+|x-3| &=(x-1)+(x-2)+(x-3)\\ &=3x-6. \end{aligned} \]
Split and evaluate the definite integral
Using the expressions obtained in the three subintervals,
\[I=\int_1^2(4-x)\,dx+\int_2^3x\,dx+\int_3^4(3x-6)\,dx\]
Evaluating the first integral,
\[ \begin{aligned} I_1 &=\int_1^2(4-x)\,dx\\ &=\left[4x-\frac{x^2}{2}\right]_1^2\\ &=\left(8-\frac{4}{2}\right) -\left(4-\frac{1}{2}\right)\\ &=6-\frac{7}{2}\\ &=\frac{5}{2}. \end{aligned} \]
Evaluating the second integral,
\[ \begin{aligned} I_2 &=\int_2^3x\,dx\\ &=\left[\frac{x^2}{2}\right]_2^3\\ &=\frac{9}{2}-\frac{4}{2}\\ &=\frac{5}{2} \end{aligned} \]
Evaluating the third integral,
\[ \begin{aligned} I_3 &=\int_3^4(3x-6)\,dx\\ &=\left[\frac{3x^2}{2}-6x\right]_3^4\\ &=(24-24)-\left(\frac{27}{2}-18\right)\\ &=0-\left(-\frac{9}{2}\right)\\ &=\frac{9}{2} \end{aligned} \]
Add the three results
\[ \begin{aligned} I&=I_1+I_2+I_3\\ &=\frac{5}{2}+\frac{5}{2}+\frac{9}{2}\\ &=\frac{19}{2}. \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int_1^4\left[|x-1|+|x-2|+|x-3|\right]dx=\frac{19}{2}}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: This problem tests the correct handling of absolute-value functions and the evaluation of definite integrals over piecewise-defined functions.
JEE Main and other entrance examinations: Recognising sign-change points and splitting the integration interval correctly is an important problem-solving technique.
Conceptual understanding: The geometrical interpretation provides an alternative way to verify the result and connects integration with area under a curve.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Always identify the zeros of the expressions inside absolute-value signs before integrating.
Divide the integration interval at every relevant sign-change point.
Remove each modulus sign according to the sign of its expression on the corresponding subinterval.
Evaluate the resulting integrals separately and add their values.
For piecewise-linear integrands, geometrical area calculations can provide a useful independent verification.

📘 Concept & Theory

Theory and Concept

›

Partial fraction decomposition is useful when the integrand is a rational function, that is, a quotient of two polynomials.

For a rational function with a repeated linear factor such as \(x^2\) and another distinct linear factor such as \(x+1\), we write:

\[ \frac{1}{x^2(x+1)} =\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1} \]

The constants \(A\), \(B\), and \(C\) are determined by comparing coefficients.

🗺️ Solution Roadmap

Step-by-step Plan

›

Express the rational function as a sum of partial fractions.
Determine the unknown constants.
Integrate each term separately.
Apply the given limits \(1\) and \(3\).
Simplify the logarithmic expression and verify the result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Given —
\[I=\int_1^3\frac{dx}{x^2(x+1)}\]
Resolve into partial fractions
Assume that
\[\frac{1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}\]
Taking the LCM and comparing numerators,
\[1=Ax(x+1)+B(x+1)+Cx^2\]
Expanding,
\[1=(A+C)x^2+(A+B)x+B\]
Comparing coefficients, we obtain
\[A+C=0,\qquad A+B=0,\qquad B=1\]
From \(B=1\),
\[A=-1,\quad C=1\]
Therefore,
\[\boxed{\frac{1}{x^2(x+1)}=-\frac1x+\frac1{x^2}+\frac1{x+1}}\]
Integrate term by term
Substituting the partial fractions,
\[I=\int_1^3\left(-\frac1x+\frac1{x^2}+\frac1{x+1}\right)dx\]
Integrating each term,
\[ \begin{aligned} I &=\left[ -\log|x|-\frac1x+\log|x+1| \right]_1^3\\ &=\left[ \log\left|\frac{x+1}{x}\right|-\frac1x \right]_1^3 \end{aligned} \]
Apply the limits
Substituting the upper limit \(x=3\),
\[\log\frac{3+1}{3}-\frac13=\log\frac43-\frac13\]
Substituting the lower limit \(x=1\),
\[\log\frac{1+1}{1}-1=\log2-1\]
Taking the difference,
\[ \begin{aligned} I &=\left(\log\frac43-\frac13\right) -(\log2-1)\\ &=\log\frac43-\log2-\frac13+1\\ &=\log\left(\frac{4}{3}\cdot\frac12\right) +\frac23\\ &=\log\frac23+\frac23 \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int_1^3\frac{dx}{x^2(x+1)}=\frac23+\log\frac23}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12: Demonstrates the use of partial fractions in evaluating definite integrals.
JEE Main: Tests decomposition involving repeated linear factors.
JEE Advanced: Reinforces algebraic manipulation and logarithmic simplification.
CUET: Provides practice in integrating rational functions and applying definite limits.

📜 Note

Important Notes

›

Since \(\frac23<1\),

\[ \log\frac23=-\log\frac32 \]

Thus, an equivalent form of the answer is

\[ \boxed{ \frac23-\log\frac32 } \]

The integral is positive because its integrand is positive throughout the interval \([1,3]\). The equivalent expression above is also positive.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Use partial fractions for rational functions with factorable denominators.
For a repeated linear factor \(x^2\), include both \(\frac{A}{x}\) and \(\frac{B}{x^2}\).
Combine logarithmic terms using logarithm properties before applying the final simplification.
Always check the sign of the logarithmic expression and the positivity of the definite integral.
The final answer is \(\boxed{\frac23+\log\frac23}\).

📘 Concept & Theory

Theory and Concept

›

This problem involves the integration by parts method, which is based on the product rule of differentiation.

The integration by parts formula is

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

For an integral involving the product of an algebraic function and an exponential function, we generally choose the algebraic function as \(u\) and the exponential function as \(dv\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Choose \(u=x\) and \(dv=e^x\,dx\).
Calculate \(du\) and \(v\).
Apply the integration by parts formula.
Evaluate the resulting definite integral using the given limits.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 5 steps

Given —
\[I=\int_0^1 xe^x\,dx\]
Choose the parts
Let
\[u=x,\qquad dv=e^x\,dx\]
Differentiating \(u\) and integrating \(dv\), we obtain
\[du=dx,\qquad v=e^x\]
Apply integration by parts
Using
\[\int u\,dv=uv-\int v\,du\]
we get
\[ \begin{aligned} I &=\left[xe^x\right]_0^1-\int_0^1e^x\,dx\\ &=\left[xe^x-e^x\right]_0^1\\ &=\left[e^x(x-1)\right]_0^1 \end{aligned} \]
Apply the limits
\[ \begin{aligned} I &=\left[e^x(x-1)\right]_0^1\\ &=e^1(1-1)-e^0(0-1)\\ &=e(0)-1(-1)\\ &=0+1\\ &=1 \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int_0^1xe^x\,dx=1}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: Tests the direct application of integration by parts to a definite integral.
JEE Main and other entrance examinations: Reinforces the selection of \(u\) and \(dv\), correct use of limits, and evaluation of exponential integrals.
Conceptual understanding: Demonstrates how integration by parts converts a product integral into a simpler integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Use integration by parts when the integrand is a product of two functions.
For \(xe^x\), choosing \(u=x\) and \(dv=e^x\,dx\) simplifies the calculation.
In definite integration, apply the limits only after obtaining the antiderivative.
Verify the antiderivative by differentiation to avoid sign and algebraic errors.

📘 Concept & Theory

Theory and Concept

›

A function \(f(x)\) is called an odd function if

\[ f(-x)=-f(x) \]

A fundamental property of definite integrals states that if \(f(x)\) is continuous and odd over a symmetric interval \([-a,a]\), then

\[ \boxed{\int_{-a}^{a}f(x)\,dx=0} \]

Similarly, a function \(g(x)\) is even if

\[ g(-x)=g(x) \]

The product of an odd function and an even function is always odd.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the parity of \(x^{17}\).
Identify the parity of \(\cos^4x\).
Determine the parity of their product.
Apply the property of definite integrals over symmetric limits.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Given —
\[I=\int_{-1}^{1}x^{17}\cos^4x\,dx\]
Check the parity of \(x^{17}\)
Replacing \(x\) by \(-x\),
\[(-x)^{17}=-x^{17}\]
Therefore, \(x^{17}\) is an odd function.
Check the parity of \(\cos^4x\)
Since cosine is an even function,
\[\cos(-x)=\cos x\]
Raising both sides to the fourth power,
\[\cos^4(-x)=\cos^4x\]
Therefore, \(\cos^4x\) is an even function.
Determine the parity of the product
Let
\[f(x)=x^{17}\cos^4x\]
Replacing \(x\) by \(-x\),
\[ \begin{aligned} f(-x) &=(-x)^{17}\cos^4(-x)\\ &=-x^{17}\cos^4x\\ &=-f(x) \end{aligned} \]
Thus, the integrand \(x^{17}\cos^4x\) is an odd function
Apply the property of definite integrals
The limits of integration are \(-1\) and \(1\), which form a symmetric interval about the origin.
Therefore, using the property of odd functions,
\[\int_{-1}^{1}f(x)\,dx=0\]
Hence,
\[I=\int_{-1}^{1}x^{17}\cos^4x\,dx=0\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int_{-1}^{1}x^{17}\cos^4x\,dx=0}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12: Tests the fundamental properties of definite integrals and even-odd functions.
JEE Main: Helps identify integrals that can be evaluated without performing lengthy calculations.
JEE Advanced: Reinforces parity analysis and symmetry-based evaluation.
CUET: Develops speed and accuracy in applying standard definite-integral properties.

📜 Note

Important Notes

›

For a symmetric interval \([-a,a]\), remember these two important properties:

For an odd function:

\[ \int_{-a}^{a}f(x)\,dx=0 \]

For an even function:

\[ \int_{-a}^{a}f(x)\,dx =2\int_0^a f(x)\,dx \]

In this problem, the product is odd because an odd function multiplied by an even function gives an odd function. The symmetric limits therefore make the integral zero.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

\(x^{17}\) is odd.
\(\cos^4x\) is even.
The product of an odd and an even function is odd.
The integral of an odd function over symmetric limits is zero.
The final answer is \(\boxed{0}\).

📘 Concept & Theory

Theory and Concept

›

This problem involves the integration of an odd power of a trigonometric function. When the power of sine is odd, we can separate one factor of \(\sin x\) and express the remaining even power in terms of cosine using the identity

\[ \sin^2x=1-\cos^2x. \]

We can then use the substitution

\[ t=\cos x,\qquad dt=-\sin x\,dx. \]

This converts the trigonometric integral into a simple polynomial integral.

🗺️ Solution Roadmap

Step-by-step Plan

›

Express \(\sin^3x\) as \(\sin x(1-\cos^2x)\).
Substitute \(t=\cos x\).
Change the limits according to the substitution.
Evaluate the resulting polynomial integral.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Given —
\[I=\int_0^{\frac{\pi}{2}}\sin^3x\,dx\]
Rewrite the integrand
Using the identity \(\sin^2x=1-\cos^2x\),
\[ \begin{aligned} I &=\int_0^{\frac{\pi}{2}}\sin x\cdot\sin^2x\,dx\\ &=\int_0^{\frac{\pi}{2}}\sin x(1-\cos^2x)\,dx \end{aligned} \]
Apply substitution
Let
\[t=\cos x\]
Differentiating,
\[dt=-\sin x\,dx\]
Therefore,
\[\sin x\,dx=-dt\]
Change the limits
When \(x=0\),
\[t=\cos 0=1\]
When \(x=\dfrac{\pi}{2}\),
\[t=\cos\frac{\pi}{2}=0\]
Substituting the new variable and limits,
\[ \begin{aligned} I &=-\int_1^0(1-t^2)\,dt\\ &=\int_0^1(1-t^2)\,dt \end{aligned} \]
Evaluate the integral
\[ \begin{aligned} I &=\left[t-\frac{t^3}{3}\right]_0^1\\ &=\left(1-\frac{1}{3}\right) -\left(0-\frac{0}{3}\right)\\ &=1-\frac{1}{3}\\ &=\frac{2}{3} \end{aligned} \]
\[\boxed{\displaystyle\int_0^{\frac{\pi}{2}}\sin^3x\,dx=\frac{2}{3}}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int_0^{\frac{\pi}{2}}\sin^3x\,dx=\frac{2}{3}}\]

🎯 Exam Significance

Exam Significance

›

CBSE Board Examinations: Tests the use of trigonometric identities and substitution in definite integration.
JEE Main and other entrance examinations: Reinforces a standard technique for evaluating integrals involving odd powers of sine.
Conceptual understanding: Highlights the importance of transforming trigonometric expressions into algebraic integrals through suitable substitutions.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For an odd power of sine, separate one factor of \(\sin x\).
Convert the remaining even power using \(\sin^2x=1-\cos^2x\).
The substitution \(t=\cos x\) simplifies the integral.
Always transform the limits when changing the variable in a definite integral.
The final value is positive, consistent with the non-negative integrand on the given interval.

📘 Concept & Theory

Theory and Concept

›

For integrals involving odd powers of tangent, we can use the identity

\[ \tan^2x=\sec^2x-1 \]

This converts the integrand into a form involving \(\sec^2x\), whose integral is straightforward using the substitution \(u=\tan x\).

We also use the standard results:

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

\[ \int\tan x\,dx=\log|\sec x|+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Express \(\tan^3x\) as \(\tan x(\sec^2x-1)\).
Split the integral into two simpler integrals.
Evaluate the resulting terms using standard integration formulas.
Apply the limits \(0\) and \(\frac{\pi}{4}\).
Simplify the logarithmic expression.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 8 steps

Given —
\[I=\int_0^{\pi/4}2\tan^3x\,dx\]
Apply the trigonometric identity
Using
\[\tan^2x=\sec^2x-1\]
we obtain
\[ \begin{aligned} I &=2\int_0^{\pi/4}\tan x(\sec^2x-1)\,dx\\ &=2\int_0^{\pi/4}\tan x\sec^2x\,dx -2\int_0^{\pi/4}\tan x\,dx \end{aligned} \]
Evaluate the first integral
Let
\[u=\tan x\]
Then,
\[du=\sec^2x\,dx\]
Therefore,
\[ \begin{aligned} 2\int_0^{\pi/4}\tan x\sec^2x\,dx &=2\int_0^1u\,du\\ &=2\left[\frac{u^2}{2}\right]_0^1\\ &=[u^2]_0^1\\ &=1 \end{aligned} \]
Here, the transformed limits are \(u=0\) when \(x=0\), and \(u=1\) when \(x=\frac{\pi}{4}\).
Evaluate the second integral
Using the standard formula,
\[\int\tan x\,dx=\log|\sec x|+C\]
we get
\[ \begin{aligned} 2\int_0^{\pi/4}\tan x\,dx &=2\left[\log|\sec x|\right]_0^{\pi/4}\\ &=2\left(\log\sqrt2-\log1\right)\\ &=2\log\sqrt2\\ &=\log2 \end{aligned} \]
Combine the results
Substituting the values of both integrals,
\[\begin{aligned}I&=1-\log2\end{aligned}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int_0^{\pi/4}2\tan^3x\,dx=1-\log2}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12: Tests the use of trigonometric identities in evaluating definite integrals.
JEE Main: Reinforces the integration of odd powers of tangent using standard identities.
JEE Advanced: Develops fluency in transforming trigonometric expressions into simpler algebraic forms.
CUET: Provides practice in combining standard integration formulas and applying definite limits.

📜 Note

Important Notes

›

For odd powers of tangent, separating one factor of \(\tan x\) and using

\[ \tan^2x=\sec^2x-1 \]

often simplifies the integration.

Also, since \(\sec x\) is positive throughout the interval \(\left[0,\frac{\pi}{4}\right]\), the absolute value signs in \(\log|\sec x|\) may be omitted.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Use \(\tan^2x=\sec^2x-1\) to simplify odd powers of tangent.
Use \(u=\tan x\) for integrals containing \(\sec^2x\,dx\).
Remember that \(\int\tan x\,dx=\log|\sec x|+C\).
Apply the limits only after obtaining the antiderivative or evaluating the transformed definite integrals.
The final answer is \(\boxed{1-\log2}\).

📘 Concept & Theory

Theory and Concept

›

Integration by parts is particularly useful for integrals involving inverse trigonometric functions.

The standard formula is

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

We also use the derivative formula

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

For this problem, we choose \(u=\sin^{-1}x\) because its derivative is simpler than the original inverse trigonometric function.

🗺️ Solution Roadmap

Step-by-step Plan

›

Choose the first and second functions for integration by parts.
Determine their derivatives and integrals.
Apply the integration-by-parts formula.
Evaluate the resulting elementary integral.
Substitute the limits \(0\) and \(1\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 15 steps

Given —
\[I=\int_0^1\sin^{-1}x\,dx\]
Choose the functions
Let
\[u=\sin^{-1}x,\quad dv=dx\]
Differentiating and integrating respectively,
\[du=\frac{dx}{\sqrt{1-x^2}},\quad v=x\]
Apply integration by parts.
Using the formula
\[\int u\,dv=uv-\int v\,du\]
we obtain
\[\begin{aligned}I&=\left[x\sin^{-1}x\right]_0^1-\int_0^1\frac{x}{\sqrt{1-x^2}}\,dx\end{aligned}\]
Evaluate the first term
Using
\[\sin^{-1}1=\frac{\pi}{2},\qquad \sin^{-1}0=0\]
we get
\[ \begin{aligned} \left[x\sin^{-1}x\right]_0^1 &=1\cdot\frac{\pi}{2}-0\cdot0\\ &=\frac{\pi}{2} \end{aligned} \]
Evaluate the remaining integral
Consider
\[J=\int_0^1\frac{x}{\sqrt{1-x^2}}\,dx\]
Put
\[t=1-x^2\]
Differentiating,
\[dt=-2x\,dx\]
Therefore,
\[x\,dx=-\frac12dt\]
Transforming the limits:
When \(x=0\), \(t=1\).
When \(x=1\), \(t=0\).
Thus,
\[ \begin{aligned} J &=-\frac12\int_1^0t^{-1/2}\,dt\\ &=\frac12\int_0^1t^{-1/2}\,dt\\ &=\frac12\left[2\sqrt t\right]_0^1\\ &=1 \end{aligned} \]
Combine the results
Since
\[I=\frac{\pi}{2}-J\]
we obtain
\[I=\frac{\pi}{2}-1\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int_0^1\sin^{-1}x\,dx=\frac{\pi}{2}-1}\]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12: Tests integration by parts involving inverse trigonometric functions.
JEE Main: Reinforces the selection of an appropriate first function and the evaluation of a standard radical integral.
JEE Advanced: Develops the ability to combine integration by parts with substitution.
CUET: Provides practice in evaluating definite integrals involving inverse trigonometric functions.

📜 Note

Important Notes

›

The derivative formula

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

is valid for \(-1 < x < 1\). Although the derivative becomes unbounded at \(x=1\), the resulting singularity is integrable. The definite integral is therefore well-defined as an improper integral at that endpoint.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Use integration by parts for integrals involving inverse trigonometric functions.
Choose \(u=\sin^{-1}x\) and \(dv=dx\).
Use substitution to evaluate the resulting radical integral.
Remember the antiderivative:

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

Q38

LONG ANSWER4 marks

Choose the correct answers \[\int \dfrac{dx}{e^{x}+e^{-x}}\] is equal to \[\begin{aligned}&\left( A\right) \quad\tan ^{-1}\left( e^{x}\right) +C\\[5pt] &\left( B\right) \quad\tan ^{-1}\left( e^{-x}\right) +C\\[5pt] &\left( C\right) \quad\log \left( e^{x}-e^{-x}\right) +C\\[5pt] &\left( D\right) \quad\log \left( e^{x}+e^{-x}\right) +C\end{aligned}\]

📘 Concept & Theory

Theory and Concept

›

We use the standard integration formula

\[ \int\frac{du}{1+u^2}=\tan^{-1}u+C \]

We also use the exponential identity

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

This allows us to transform the given integral into the standard inverse tangent form.

🗺️ Solution Roadmap

Step-by-step Plan

›

Multiply the numerator and denominator by \(e^x\).
Simplify the resulting expression.
Use the substitution \(u=e^x\).
Apply the standard inverse tangent integration formula.
Compare the result with the given options.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 6 steps

Given —
\[I=\int\frac{dx}{e^x+e^{-x}}\]
Multiply the numerator and denominator by \(e^x\).
\[ \begin{aligned} I &=\int\frac{e^x\,dx}{e^{2x}+1} \end{aligned} \]
Apply substitution
Let
\[u=e^x\]
Differentiating,
\[du=e^x\,dx\]
Therefore,
\[I=\int\frac{du}{1+u^2}\]
Apply the standard formula
\[I=\tan^{-1}u+C\]
Substituting \(u=e^x\),
\[\boxed{I=\tan^{-1}(e^x)+C}\]

💡 Answer

Final Answer

›

Comparing the result with the given options:

(A) \(\tan^{-1}(e^x)+C\) — Correct
(B) \(\tan^{-1}(e^{-x})+C\) — Incorrect
(C) \(\log(e^x-e^{-x})+C\) — Incorrect
(D) \(\log(e^x+e^{-x})+C\) — Incorrect

\[ \boxed{\text{Correct option: (A)}} \]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12: Tests substitution and standard inverse trigonometric integration formulas.
JEE Main: Reinforces exponential manipulation and recognition of standard integral forms.
JEE Advanced: Develops the ability to identify suitable substitutions in exponential-rational expressions.
CUET: Tests quick identification of correct antiderivatives through differentiation.

📜 Note

Important Notes

›

Option (B) may appear similar, but differentiating it gives

\[ \begin{aligned} \frac{d}{dx}\tan^{-1}(e^{-x}) &=-\frac{e^{-x}}{1+e^{-2x}}\\ &=-\frac{1}{e^x+e^{-x}} \end{aligned} \]

It has the opposite sign of the given integrand. In fact,

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

Thus, option (B) is not an antiderivative of the given function.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Multiply the numerator and denominator by \(e^x\) to simplify the expression.
Use \(u=e^x\), so that \(du=e^x\,dx\).
Recognise the standard form \(\int\frac{du}{1+u^2}\).
Verify multiple-choice integration answers by differentiation.
The correct answer is \(\boxed{\text{Option (A)}}\).

Q39

LONG ANSWER4 marks

Choose the correct answer \[\int \dfrac{\cos 2x}{\left( \sin x+\cos x\right) ^{2}}dx\] is equal to \[\begin{aligned}&\left( A\right) \quad\dfrac{-1}{\sin x+\cos x}+C\\[5pt] &\left( B\right) \quad\log \left| \sin x+\cos x\right| +C\\[5pt] &\left( C\right) \quad\log \left| \sin x-\cos x\right| +C\\[5pt] &\left( D\right) \quad\dfrac{1}{\left( \sin x+\cos x\right) ^{2}}\end{aligned}\]

📘 Concept & Theory

Theory and Concept

›

This problem uses the trigonometric identity

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

By factorisation,

\[ \cos 2x=(\cos x-\sin x)(\cos x+\sin x) \]

We also use the standard integration formula

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

🗺️ Solution Roadmap

Step-by-step Plan

›

Factorise the numerator using the identity for \(\cos 2x\).
Cancel the common factor \(\sin x+\cos x\).
Recognise the resulting expression as the logarithmic derivative of \(\sin x+\cos x\).
Integrate and compare the result with the given options.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 8 steps

Given —
\[I=\int\frac{\cos 2x}{(\sin x+\cos x)^2}\,dx\]
Use the trigonometric identity
We know that
\[\cos 2x=\cos^2x-\sin^2x\]
Applying the difference of two squares,
\[\cos 2x=(\cos x-\sin x)(\cos x+\sin x)\]
Therefore,
\[I=\int\frac{(\cos x-\sin x)(\cos x+\sin x)}{(\sin x+\cos x)^2}\,dx\]
Simplify the integrand
Cancelling one common factor,
\[I=\int\frac{\cos x-\sin x}{\sin x+\cos x}\,dx\]
Apply substitution
Let
\[u=\sin x+\cos x\]
Differentiating,
\[du=(\cos x-\sin x)\,dx\]
Hence,
\[ \begin{aligned} I&=\int\frac{du}{u}\\ &=\log|u|+C \end{aligned} \]
Substituting \(u=\sin x+\cos x\), we obtain
\[\boxed{I=\log|\sin x+\cos x|+C}\]

💡 Answer

Final Answer

›

Comparing the result with the given options:

(A) \(\dfrac{-1}{\sin x+\cos x}+C\) — Incorrect
(B) \(\log|\sin x+\cos x|+C\) — Correct
(C) \(\log|\sin x-\cos x|+C\) — Incorrect
(D) \(\dfrac{1}{(\sin x+\cos x)^2}\) — Incorrect

\[ \boxed{\text{Correct option: (B)}} \]

🎯 Exam Significance

Exam Significance

›

CBSE Class 12: Tests the application of trigonometric identities and logarithmic integration.
JEE Main: Develops the ability to simplify trigonometric expressions before integration.
JEE Advanced: Reinforces recognition of logarithmic derivatives hidden within algebraic-trigonometric expressions.
CUET: Tests quick identification of standard integration patterns.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Use \(\cos 2x=(\cos x-\sin x)(\cos x+\sin x)\).
Simplify the integrand before selecting a substitution.
Recognise the logarithmic derivative of \(\sin x+\cos x\).
Include the absolute value in the logarithmic result.
The correct answer is \(\boxed{\text{Option (B)}}\).

Q40

LONG ANSWER4 marks

If \[f\left( a+b-x\right) =f\left( x\right) \] then \[\int ^{b}_{a}xf\left( x\right) dx\] is equal to \[\begin{aligned}&\left( A\right) \quad\dfrac{a+b}{2}\int ^{b}_{a}f\left( b-x\right) dx\\[5pt] &\left( B\right) \quad\dfrac{a+b}{2}\int ^{b}_{a}f\left( b+x\right) dx\\[5pt] &\left( C\right) \quad\dfrac{b-a}{2}\int ^{b}_{a}f\left( x\right) dx_{b}\\[5pt] &\left( D\right) \quad\dfrac{a+b}{2}\int _{a}f\left( x\right) dx\end{aligned}\]

📘 Concept & Theory

Theory and Concept

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We use the substitution property of definite integrals:

\[ \int_a^b g(x)\,dx = \int_a^b g(a+b-x)\,dx \]

We are also given the symmetry condition

\[ f(a+b-x)=f(x) \]

This means that the function is symmetric about the midpoint \(x=\dfrac{a+b}{2}\) of the interval \([a,b]\).

🗺️ Solution Roadmap

Step-by-step Plan

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Let the given integral be \(I\).
Apply the substitution \(x=a+b-t\).
Use the given symmetry condition to simplify the transformed integral.
Add the original and transformed expressions.
Compare the resulting identity with the given options.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Given —
\[I=\int_a^b xf(x)\,dx\]
Apply substitution
Let
\[x=a+b-t\]
Then
\[dx=-dt\]
Changing the limits:
\[ \begin{aligned} x=a&\implies t=b\\ x=b&\implies t=a \end{aligned} \]
Therefore,
\[ \begin{aligned} I &=\int_b^a(a+b-t)f(a+b-t)(-dt)\\ &=\int_a^b(a+b-t)f(a+b-t)\,dt \end{aligned} \]
Use the given condition
Since
\[f(a+b-t)=f(t)\]
we obtain
\[I=\int_a^b(a+b-t)f(t)\,dt\]
Separating the terms,
\[I=(a+b)\int_a^b f(t)\,dt-\int_a^b tf(t)\,dt\]
But
\[\int_a^b tf(t)\,dt=I\]
Hence,
\[I=(a+b)\int_a^b f(t)\,dt-I\]
Simplify
Rearranging,
\[2I=(a+b)\int_a^b f(t)\,dt\]
Therefore,
\[\boxed{I=\frac{a+b}{2}\int_a^b f(x)\,dx}\]
Since the given symmetry condition also implies \(f(b-x)=f(x-a)\) only under an additional shift relation, the expression in option (A) is not generally the same as the derived integral. We must compare the options carefully.

💡 Answer

Final Answer

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The correct identity is

\[ \boxed{ \int_a^b xf(x)\,dx = \frac{a+b}{2}\int_a^b f(x)\,dx } \]

Thus, the intended answer is option (D) if its integral is understood as \(\int_a^b f(x)\,dx\).

Note: Option (D) in the question appears to have a missing upper limit \(b\). Option (A), as printed, contains \(\int_a^b f(b-x)\,dx\), which is not generally equivalent to \(\int_a^b f(x)\,dx\) for arbitrary \(a,b\).

🎯 Exam Significance

Exam Significance

›

CBSE Class 12: Tests the symmetry property of definite integrals.
JEE Main: Requires efficient use of the substitution \(x=a+b-t\).
JEE Advanced: Develops the use of functional symmetry to simplify definite integrals.
CUET: Tests recognition of standard definite-integral identities.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Use the substitution \(x=a+b-t\) for symmetric functions.
Change both the limits and the differential carefully.
Apply the given symmetry condition after substitution.
Add the original and transformed integrals to obtain the identity.
The mathematically correct expression is \(\boxed{\text{Option (D), as intended}}\), provided the missing upper limit \(b\) is restored.

Integrals

Comparing the constant terms

Comparing the coefficients of \(x\):

Comparing the coefficients of \(x^2\):

Chapter Complete!

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APPLICATION OF DERIVATIVES — Learning Resources

Frequently Asked Questions

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Integrals

Comparing the constant terms

Comparing the coefficients of \(x\):

Comparing the coefficients of \(x^2\):

Chapter Complete!

Recent posts

APPLICATION OF DERIVATIVES — Learning Resources

Frequently Asked Questions

Q 01 What is the symmetry property of definite integrals used in Question 40?

Q 02 What is the given condition in NCERT Class 12 Maths Integrals Exercise 7.10 Question 40?

Q 03 Which substitution is used to solve Question 40?

Q 04 What is the value of the integral ??? xf(x) dx under the given symmetry condition?

Q 05 How is the symmetry condition applied after substitution?

Q 06 Why are the limits interchanged during the substitution x=a+b-t?

Q 07 Which option represents the mathematically correct identity in Question 40?

Q 08 Why is option (A) not generally equivalent to the derived result?

Q 09 What is the geometric significance of the symmetry property?

Q 10 How is the symmetry property of definite integrals useful in competitive examinations?

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