# JEE Main Indefinite Integrals Previous Year Questions With Solutions

JEE Previous year questions on Indefinite integrals gives students the opportunity to learn the right method of solving questions related to important concepts like indefinite integral, integration using partial fractions and integration by parts. The solutions provided here will serve as a great reference tool for students who are preparing for JEE exams. About 2-4 questions are asked from this topic in JEE Examination. Students are advised to download the set of past year important questions which will help them to go through and revise solutions quickly before the exam.

## JEE Main Indefinite Integration Past Year Questions With Solutions

Question 1: Evaluate $\int_{{}}^{{}}{\frac{x-1}{{{(x+1)}^{3}}}{{e}^{x}}\ dx}$

Solution:

$\int \frac{x-1}{(x+1)^3} \; e^x dx$ $=\int_{{}}^{{}}{{{e}^{x}}\left( \frac{(x+1)}{{{(x+1)}^{3}}}-\frac{2}{{{(x+1)}^{3}}} \right)\,dx}$ $=\int_{{}}^{{}}{{{e}^{x}}\left( \frac{1}{{{(x+1)}^{2}}}-\frac{2}{{{(x+1)}^{3}}} \right)\,dx}$ $=\frac{{{e}^{x}}}{{{(x+1)}^{2}}}+c$

Using formula, $\frac{d}{dx}\left( \frac{1}{{{(x+1)}^{2}}} \right) =- \frac{2}{(x+1)^3}$

Question 2: If $\int_{{}}^{{}}{\frac{{{e}^{x}}(1+\sin x)dx}{1+\cos x}={{e}^{x}}f(x)+c},$ then find $f(x)$.

Solution:

$I=\int_{{}}^{{}}{{{e}^{x}}\left( \frac{1+\sin x}{1+\cos x} \right)\,dx}=\int_{{}}^{{}}{{{e}^{x}}\left[ \frac{1+2\sin (x/2)\,\cos (x/2)}{2{{\cos }^{2}}(x/2)} \right]dx}\\ I=\int_{{}}^{{}}{{{e}^{x}}\left[ \frac{1}{2}{{\sec }^{2}}(x/2)+\tan (x/2) \right]\,dx}={{e}^{x}}.\tan (x/2)+c$

Because, $\int_{{}}^{{}}{{{e}^{x}}[f(x)+{f}'(x)\,]dx={{e}^{x}}.\,f(x)+c\}}$

Question 3: Evaluate $\int_{{}}^{{}}{x{{\sin }^{2}}x\ dx}$

Solution:

$\int_{{}}^{{}}{x{{\sin }^{2}}x\,dx}=\int_{{}}^{{}}{x\,.\,\frac{(1-\cos 2x)}{2}\,dx}\\ =\frac{1}{2}\left[ \int_{{}}^{{}}{x\,dx}-\int_{{}}^{{}}{x\,.\,\cos 2x\,dx} \right]\\ =\frac{{{x}^{2}}}{4}-\frac{x}{4}\sin 2x-\frac{1}{8}\cos 2x+c.$

Question 4: Evaluate $\int_{{}}^{{}}{{{x}^{2}}\sin 2x}\ dx$

Solution:

$I=\int_{{}}^{{}}{{{x}^{2}}\sin 2x\,dx}=\frac{-{{x}^{2}}\cos 2x}{2}+\int_{{}}^{{}}{\frac{2x\cos 2x}{2}\,dx}+c \\ =-\frac{{{x}^{2}}\cos 2x}{2}+\frac{x\sin 2x}{2}+\frac{\cos 2x}{4}+c.$

Question 5: Evaluate $\int_{{}}^{{}}{{{e}^{x}}\sin x\ dx}$

Solution:

$I=\int_{{}}^{{}}{{{e}^{x}}\sin x\,dx}={{e}^{x}}\sin x-\int_{{}}^{{}}{{{e}^{x}}\cos x\,dx+c}\\ ={{e}^{x}}\sin x-{{e}^{x}}\cos x-\int_{{}}^{{}}{{{e}^{x}}\sin x\,dx+c}\\ 2I={{e}^{x}}(\sin x-\cos x)+c\\ I=\frac{1}{2}{{e}^{x}}(\sin x-\cos x)+c.$

Question 6: Evaluate $\int_{{}}^{{}}{\cos \sqrt{x}\ dx}$

Solution:

Put $\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}\,dx=dt\Rightarrow dx=2t\,dt,$

It reduces to

$\int_{{}}^{{}}{2t\,.\cos t\,dt}=2\left[ t\,.\,\sin t-\int_{{}}^{{}}{\sin t\,dt} \right]\\ =2t\sin t+2\cos t\\ =2[\sqrt{x}\sin \sqrt{x}+\cos \sqrt{x}]+c.$

Question 7: Simplify $\int_{{}}^{{}}{\frac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\ }dx$

Solution:

Putting ${{\sin }^{-1}}x=t\Rightarrow \frac{1}{\sqrt{1-{{x}^{2}}}}\,dx=dt,$

We get

$\int_{{}}^{{}}{\frac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\,dx=}\int_{{}}^{{}}{t\sin t\,dt=-t\cos t+\sin t+c}\\ =-{{\sin }^{-1}}x\cos ({{\sin }^{-1}}x)+\sin ({{\sin }^{-1}}x)+c\\ =x-{{\sin }^{-1}}x\sqrt{1-{{x}^{2}}}+c.$

Question 8: Evaluate $\int_{{}}^{{}}{\frac{dx}{\sin x+\cos x}}$

Solution:

$\int_{{}}^{{}}{\frac{dx}{\sin x+\cos x}}=\frac{1}{\sqrt{2}}\int_{{}}^{{}}{\frac{dx}{\sin x\cos \frac{\pi }{4}+\cos x\sin \frac{\pi }{4}}} \\ =\frac{1}{\sqrt{2}}\int_{{}}^{{}}{\text{cosec }\left( x+\frac{\pi }{4} \right)\,dx=\frac{1}{\sqrt{2}}\log \tan \left( \frac{\pi }{8}+\frac{x}{2} \right)}+c.$

Question 9: Evaluate $\int{\sqrt{{{x}^{2}}+{{a}^{2}}}\,\,dx}$

Solution:

$I=\int{\sqrt{{{x}^{2}}+{{a}^{2}}}\,dx}\\ =\int{\sqrt{{{x}^{2}}+{{a}^{2}}}\,.1\,dx}\\ =\sqrt{{{x}^{2}}+{{a}^{2}}}\int{1dx-\int{\left[ \frac{d}{dx}\left( \sqrt{{{x}^{2}}+{{a}^{2}}} \right)\int{1\,dx} \right]\,dx}}\\ =x\sqrt{{{x}^{2}}+{{a}^{2}}}-\int{\left[ \frac{2x}{2\sqrt{{{x}^{2}}+{{a}^{2}}}}x \right]}\,dx\\ =x\sqrt{{{x}^{2}}+{{a}^{2}}}-\int{\left[ \frac{{{x}^{2}}+{{a}^{2}}-{{a}^{2}}}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right]}dx\\ =x\sqrt{{{x}^{2}}+{{a}^{2}}}-\int{\left[ \sqrt{{{x}^{2}}+{{a}^{2}}}-\frac{{{a}^{2}}}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right]\,}dx \\ =x\sqrt{{{x}^{2}}+{{a}^{2}}}-\int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx+{{a}^{2}}\int{\frac{dx}{\sqrt{{{x}^{2}}+{{a}^{2}}}}}} \\ =\sqrt{{{x}^{2}}+{{a}^{2}}}-I+{{a}^{2}}\log \left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]+C\\ 2I=x\sqrt{{{x}^{2}}+{{a}^{2}}+}{{a}^{2}}\log \left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]+C\\ I=\frac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\frac{{{a}^{2}}}{2}\log \left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]+C.$

Question 10: Evaluate $\int{\frac{{{(x+1)}^{2}}\,\,dx}{x({{x}^{2}}+1)}}$

Solution:

$\int{\frac{{{(x+1)}^{2}}\,\,dx}{x({{x}^{2}}+1)}}$ $=\int{\frac{{{x}^{2}}+1+2x}{x({{x}^{2}}+1)}dx}$ $=\int{\frac{{{x}^{2}}+1}{x({{x}^{2}}+1)}dx+2\int{\frac{x}{x({{x}^{2}}+1)}dx}}$

$=\int{\frac{dx}{x}+2\int{\frac{dx}{{{x}^{2}}+1}}}={{\log }_{e}}x+2{{\tan }^{-1}}x+c$

Question 11: Find the value of A and B if $\int \frac{sinx}{sin(x – \alpha)} \; \; dx = Ax + B \; log \; sin(x – \alpha) + C$

Choose the right option:

(a) (sin α, cos α)                    (b) (cos α, sin α)

(c) (- sinα, cosα)                   (d) (-cosα, sinα)

Solution: Correct option is (b)

Explanation:

Let x – α = m

=> $\int \frac{sin (\alpha + m)}{sin \; m} \; \; dm = sin \alpha \int cot \; m \; dm + cos \alpha \int dm$

= cos α(x – α) + sin α ln|sin m| + c

Therefore, on comparing we have

A = cos α and B = sin α

Question 12: Solve $\int (\frac{(log \; x – 1)}{1 + (log \; x)^2})^2$ dx

Solution:

$\int (\frac{(log \; x – 1)}{1 + (log \; x)^2})^2$ dx

= $\int \frac{(log \; x – 1)^2}{(1 + (log \; x)^2)^2}$ dx

= $\int[\frac{1}{1+(logx)^2} – \frac{2logx}{1+(logx)^2}]$

Put log x = t so dx = e^t dt

=> $\int[\frac{e^t}{1+t^2} – \frac{2te^t}{(1+t^2)^2}] \; dt$

=> $\int e^t[\frac{1}{1+t^2} – \frac{2t}{(1+t^2)^2}] \; dt$

= $\frac{e^t}{1+t^2} + c$

= $\frac{x}{1+(log \; x)^2} + c$

Question 13: Find the value of $\sqrt{2} \int \frac{sin \; x}{sin(x – \pi/4)}\; dx$

Solution:

$\sqrt{2} \int \frac{sin \; x}{sin(x – \pi/4)}\; dx$

= $\sqrt{2} \int \frac{sin \; (x – \pi/4 + \pi/4)}{sin(x – \pi/4)}\; dx$

= $\sqrt{2} \int [cos(\pi/4) + cot(x – \pi/4)sin \pi/4]\; dx$

= $\int dx + \int cot(x – \pi/4) \; dx$

= x + ln|sin(x – π/4)| + c