Indefinite Integration JEE Main previous year questions with solutions

5 solved JEE Main questions on Indefinite Integration, free to read — no sign-in needed. The full chapter has 106 questions; sign in to attempt the remaining 101 in the exam simulator.

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1. Q1JEE Main 2026 (28 Jan, Shift 1)Definite Integration by Parts
   If $\int\left(\frac{1-5 \cos ^{2} x}{\sin ^{5} x \cos ^{2} x}\right) d x=f(x)+\mathrm{C}$, where C is the constant of integration, then $f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)$ is equal to
   1. A.$\frac{1}{\sqrt{3}}(26+\sqrt{3})$
   2. B.$\frac{1}{\sqrt{3}}(26-\sqrt{3})$
   3. C.$\frac{4}{\sqrt{3}}(8-\sqrt{6})$
   4. D.$\frac{2}{\sqrt{3}}(4+\sqrt{6})$
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   Answer: (C)

   $\int \frac{1 - 5\cos^2 x}{\sin^5 x \cos^2 x} dx = \int \frac{dx}{\sin^5 x \cos^2 x} - 5\int \frac{dx}{\sin^5 x}$ $= \int \frac{\sec^2 x \, dx}{\sin^5 x} - 5\int \frac{dx}{\sin^5 x}$ By IBP: $= \frac{\tan x}{\sin^5 x} - \int \left(-\frac{5}{\sin^6 x}\right) \cos x \cdot \tan x \, dx - 5\int \frac{dx}{\sin^5 x}$ $= \frac{\tan x}{\sin^5 x} + C$ So $f(x) = \frac{\tan x}{\sin^5 x}$ $f\left(\frac{\pi}{6}\right) - f\left(\frac{\pi}{4}\right) = \frac{2^5}{\sqrt{3}} - (\sqrt{2})^5 = \frac{32}{\sqrt{3}} - 4\sqrt{2} = \frac{4}{\sqrt{3}}(8 - \sqrt{6})$
2. Q2JEE Main 2025 (23 Jan, Shift 2)Integration by Parts
   <p>Let $\int x^3 \sin x \mathrm{~d} x=g(x)+C$, where $C$ is the constant of integration. If $8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)=\alpha \pi^3+\beta \pi^2+\gamma, \alpha, \beta, \gamma \in Z$, then $\alpha+\beta-\gamma$ equals :</p>
   1. A.48
   2. B.55
   3. C.62
   4. D.<p>47</p>
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   Answer: (B)

   <p>$\begin{aligned} &amp; \int x^3 \sin x d x=-x^3 \cos x+\int 3 x^2 \cos x d x \\ &amp; =-x^3 \cos x+3 x^2 \sin x-\int 6 x \sin x d x \\ &amp; =-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin x+c \end{aligned}$ So $g(x)=-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin x$ $\begin{aligned} &amp; \mathrm{g}\left(\frac{\pi}{2}\right)=\frac{3 \pi^2}{4}-6 \\ &amp; \mathrm{~g}^{\prime}(\mathrm{x})=-3 \mathrm{x}^2 \cos \mathrm{x}+\mathrm{x}^3 \sin \mathrm{x}+6 \cos \mathrm{x}-6 \cos \mathrm{x} \\ &amp; \mathrm{~g}^{\prime}\left(\frac{\pi}{2}\right)=\frac{\pi^3}{8} \\ &amp; 8\left(\mathrm{~g}\left(\frac{\pi}{2}\right)+\mathrm{g}^{\prime}\left(\frac{\pi}{2}\right)\right)=\pi^3+6 \pi^2-48 \end{aligned}$ So $\alpha+\beta-\gamma=55$</p>
3. Q3JEE Main 2024 (29 Jan, Shift 1)Definite Integration by Substitution
   For $x\in \left(-\dfrac{\pi }{2},\dfrac{\pi }{2}\right)$, if $y\left(x\right)=\int \dfrac{cosecx+\sin x}{cosecx\sec x+\tan x{\sin }^{2}x}dx$ and $\lim _{x\rightarrow {\left(\dfrac{\pi }{2}\right)}^{-}}y\left(x\right)=0$ then $y\left(\dfrac{\pi }{4}\right)$ is equal to
   1. A.${\tan }^{-1}\left(\dfrac{1}{\sqrt{2}}\right)$
   2. B.$\dfrac{1}{2}{\tan }^{-1}\left(\dfrac{1}{\sqrt{2}}\right)$
   3. C.$-\dfrac{1}{\sqrt{2}}{\tan }^{-1}\left(\dfrac{1}{\sqrt{2}}\right)$
   4. D.$\dfrac{1}{\sqrt{2}}{\tan }^{-1}\left(-\dfrac{1}{2}\right)$
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   Answer: (D)

   Given: $f\left(x\right)=\int \dfrac{cosecx+\sin x}{cosecx\sec x+\tan x{\sin }^{2}x}dx$ $\Rightarrow f\left(x\right)=\int \dfrac{\dfrac{1}{\sin x}+\sin x}{\dfrac{1}{\sin x\cos x}+\dfrac{\sin x}{\cos x}\left({\sin }^{2}x\right)}dx$ $\Rightarrow f\left(x\right)=\int \dfrac{\dfrac{1+{\sin }^{2}x}{\sin x}}{\dfrac{1+{\sin }^{4}x}{\sin x\cos x}}dx$ $\Rightarrow f\left(x\right)=\int \dfrac{\left(1+{\sin }^{2}x\right)\cos x}{1+{\sin }^{4}x}dx$ Let, $\sin x=t$ $\cos xdx=dt$ $\Rightarrow f\left(x\right)=\int \dfrac{1+{t}^{2}}{1+{t}^{4}}dt$ $\Rightarrow f\left(x\right)=\int \dfrac{1+\dfrac{1}{{t}^{2}}}{{t}^{2}+\dfrac{1}{{t}^{2}}}dt$ $\Rightarrow f\left(x\right)=\int \dfrac{1+\dfrac{1}{{t}^{2}}}{{\left(t-\dfrac{1}{t}\right)}^{2}+2}dt$ Let, $t-\dfrac{1}{t}=u$ $\Rightarrow f\left(x\right)=\int \dfrac{1}{{u}^{2}+2}du$ $\Rightarrow f\left(x\right)=\dfrac{1}{\sqrt{2}}{\tan }^{-1}\left(\dfrac{u}{\sqrt{2}}\right)+C$ $\Rightarrow f\left(x\right)=\dfrac{1}{\sqrt{2}}{\tan }^{-1}\left(\dfrac{t-\dfrac{1}{t}}{\sqrt{2}}\right)+C$ $\Rightarrow f\left(x\right)=\dfrac{1}{\sqrt{2}}{\tan }^{-1}\left(\dfrac{\sin x-\dfrac{1}{\sin x}}{\sqrt{2}}\right)+C$ $\Rightarrow f\left(x\right)=\dfrac{1}{\sqrt{2}}{\tan }^{-1}\left(\dfrac{\sin x-cosecx}{\sqrt{2}}\right)+C$ Now, $\lim _{x\rightarrow {\dfrac{\pi }{2}}^{-}}f\left(x\right)=0$ $\Rightarrow \lim _{x\rightarrow {\dfrac{\pi }{2}}^{-}}f\left(x\right)=\lim _{x\rightarrow {\dfrac{\pi }{2}}^{-}}\left[\dfrac{1}{\sqrt{2}}{\tan }^{-1}\left(\dfrac{\sin x-cosecx}{\sqrt{2}}\right)+C\right]=0$ $\Rightarrow \dfrac{1}{\sqrt{2}}{\tan }^{-1}\left(0\right)+C=0$ $\Rightarrow C=0$ $\Rightarrow f\left(x\right)=\dfrac{1}{\sqrt{2}}{\tan }^{-1}\left(\dfrac{\sin x-cosecx}{\sqrt{2}}\right)$ $\Rightarrow y\left(\dfrac{\pi }{4}\right)=\dfrac{1}{\sqrt{2}}{\tan }^{-1}\left(\dfrac{\sin \dfrac{\pi }{4}-cosec\dfrac{\pi }{4}}{\sqrt{2}}\right)$ $\Rightarrow y\left(\dfrac{\pi }{4}\right)=\dfrac{1}{\sqrt{2}}{\tan }^{-1}\left(\dfrac{\dfrac{1}{\sqrt{2}}-\sqrt{2}}{\sqrt{2}}\right)$ $\Rightarrow y\left(\dfrac{\pi }{4}\right)=\dfrac{1}{\sqrt{2}}{\tan }^{-1}\left(\dfrac{\dfrac{-1}{\sqrt{2}}}{\sqrt{2}}\right)$ $\Rightarrow y\left(\dfrac{\pi }{4}\right)=\dfrac{1}{\sqrt{2}}{\tan }^{-1}\left(\dfrac{-1}{2}\right)$
4. Q4JEE Main 2023 (12 Apr, Shift 1)Integration by Substitution
   Let $I\left(x\right)=\int \sqrt{\dfrac{x+7}{x}}dx$ and $I\left(9\right)=12+7{\log }_{e}7$. If $I\left(1\right)=\alpha +7{\log }_{e}\left(1+2\sqrt{2}\right),$ then ${\alpha }^{4}$ is equal to _____.
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   Answer: 64

   Given, $I\left(x\right)=\int \sqrt{\dfrac{x+7}{x}}dx$ Now let, $\dfrac{x+7}{x}={t}^{2}\Rightarrow -\dfrac{7}{{x}^{2}}dx=2tdt$ $\Rightarrow dx=\dfrac{-14t}{{\left({t}^{2}-1\right)}^{2}}dt$ So, $I\left(x\right)=-14\int \dfrac{{t}^{2}}{{\left({t}^{2}-1\right)}^{2}}dt$ $\Rightarrow I\left(x\right)=-14\int \dfrac{dt}{\left({t}^{2}+\dfrac{1}{{t}^{2}}-2\right)}$ $\Rightarrow I\left(x\right)=\dfrac{-14}{2}\int \left[\dfrac{\left(1-\dfrac{1}{{t}^{2}}\right)}{{\left(t+\dfrac{1}{t}\right)}^{2}-4}+\dfrac{\left(1+\dfrac{1}{{t}^{2}}\right)}{{\left(t-\dfrac{1}{t}\right)}^{2}}\right]dt$ $\Rightarrow I\left(x\right)=-7\left(\dfrac{1}{4}\ln \left|\dfrac{t+\dfrac{1}{t}-2}{t+\dfrac{1}{t}+2}\right|-\dfrac{1}{t-\dfrac{1}{t}}\right)+c$ Now when $x=9,t=\dfrac{4}{3}$ $\Rightarrow I\left(9\right)=12+7\times \ln 7=\dfrac{-7}{4}\ln {\left(\dfrac{1}{7}\right)}^{2}+7\times \dfrac{12}{7}+c$ $\Rightarrow c=\dfrac{7}{2}\ln 7$ Now when $x=1,t=2\sqrt{2}$ $\Rightarrow I\left(1\right)=+\dfrac{7}{4}\ln {\left(\dfrac{2\sqrt{2}+1}{2\sqrt{2}-1}\right)}^{2}+7\times \dfrac{2\sqrt{2}}{7}+\dfrac{7}{2}\ln 7$ $\Rightarrow I\left(1\right)=\dfrac{7}{2}\ln \left(\dfrac{{\left(2\sqrt{2}+1\right)}^{2}}{7}\right)+2\sqrt{2}+\dfrac{7}{2}\ln 7$ $\Rightarrow I\left(1\right)=7\ln \left(2\sqrt{2}+1\right)-\dfrac{7}{2}\ln 7+2\sqrt{2}+\dfrac{7}{2}\ln 7$ $\Rightarrow \alpha =2\sqrt{2}\Rightarrow {\alpha }^{4}=64$
5. Q5JEE Main 2020 (09 Jan, Shift 1)Variable Separable Form
   If ${f}^{'}\left(x\right)={\tan }^{-1}⁡\left(\sec ⁡x+\tan ⁡x\right),-\dfrac{\pi }{2}\lt x\lt \dfrac{\pi }{2}$ and $f\left(0\right)=0$ , then $f\left(1\right)$ is equal to:
   1. A.$\dfrac{\pi +1}{4}$
   2. B.$\dfrac{1}{4}$
   3. C.$\dfrac{\pi -1}{4}$
   4. D.$\dfrac{\pi +2}{4}$
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   Answer: (A)

   ${f}^{'}\left(x\right)={\tan }^{-1}⁡\left(\sec ⁡x+\tan ⁡x\right)={\tan }^{-1}⁡\left(\dfrac{1+\sin ⁡x}{\cos ⁡x}\right)={\tan }^{-1}⁡\left(\dfrac{1-\cos ⁡\left(\dfrac{\pi }{2}+x\right)}{\sin ⁡\left(\dfrac{\pi }{2}+x\right)}\right)$ $\Rightarrow {f}^{'}(x)={\tan }^{-1}⁡\left(\dfrac{2{\sin }^{2}⁡\left(\dfrac{\pi }{4}+\dfrac{x}{2}\right)}{2\sin ⁡\left(\dfrac{\pi }{4}+\dfrac{x}{2}\right)\cos ⁡\left(\dfrac{\pi }{4}+\dfrac{x}{2}\right)}\right)$ $\Rightarrow \left({f}^{'}\left(x\right)\right)dx=\left(\dfrac{\pi }{4}+\dfrac{x}{2}\right)dx$ $\Rightarrow f\left(x\right)=\dfrac{\pi }{4}x+\dfrac{{x}^{2}}{4}+c$ $∵f\left(0\right)=0\Rightarrow c=0$ So, $f\left(1\right)=\dfrac{\pi +1}{4}$