Definite Integration JEE Main previous year questions with solutions

5 solved JEE Main questions on Definite Integration, free to read — no sign-in needed. The full chapter has 301 questions; sign in to attempt the remaining 296 in the exam simulator.

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1. Q1JEE Main 2026 (04 Apr, Shift 2)Definite Integration by Parts
   The integral $\int_{0}^{1}\cot^{-1}(1+x+x^2)dx$ is equal to:
   1. A.$2\tan^{-1}2+\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)+\dfrac{\pi}{2}$
   2. B.$2\tan^{-1}2+\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)-\dfrac{\pi}{2}$
   3. C.$2\tan^{-1}2-\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)+\dfrac{\pi}{2}$
   4. D.$2\tan^{-1}2-\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)-\dfrac{\pi}{2}$
   Show answer & solution

   Answer: (D)

   $I = \int_{0}^{1}\cot^{-1}(1+x+x^2)dx$ $I = \int_{0}^{1}\tan^{-1}\left(\dfrac{1}{1+x+x^2}\right)dx$ $I = \int_{0}^{1}\tan^{-1}\left(\dfrac{(x+1)-x}{1+(x+1)x}\right)dx$ $I = \int_{0}^{1}(\tan^{-1}(x+1)-\tan^{-1}x)dx$ $I = \int_{0}^{1}\tan^{-1}(x+1)dx - \int_{0}^{1}\tan^{-1}x dx$ Substituting $x+1 = t$ in the first integral: $I = \int_{1}^{2}\tan^{-1}x dx - \int_{0}^{1}\tan^{-1}x dx$ Using integration by parts, $\int \tan^{-1}x dx = x\tan^{-1}x - \dfrac{1}{2}\log_e(1+x^2)$ $I = \left[x\tan^{-1}x - \dfrac{1}{2}\log_e(1+x^2)\right]_{1}^{2} - \left[x\tan^{-1}x - \dfrac{1}{2}\log_e(1+x^2)\right]_{0}^{1}$ $I = \left(2\tan^{-1}2 - \dfrac{1}{2}\log_e 5 - \left(\dfrac{\pi}{4} - \dfrac{1}{2}\log_e 2\right)\right) - \left(\dfrac{\pi}{4} - \dfrac{1}{2}\log_e 2 - 0\right)$ $I = 2\tan^{-1}2 - \dfrac{1}{2}\log_e 5 - \dfrac{\pi}{2} + \log_e 2$ $I = 2\tan^{-1}2 - \dfrac{1}{2}\log_e 5 + \dfrac{1}{2}\log_e 4 - \dfrac{\pi}{2}$ $I = 2\tan^{-1}2 - \dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right) - \dfrac{\pi}{2}$ Answer: $2\tan^{-1}2-\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)-\dfrac{\pi}{2}$
2. Q2JEE Main 2025 (29 Jan, Shift 1)Definite Integration by Substitution
   The integral $80 \int_0^{\frac{\pi}{4}}\left(\frac{\sin \theta+\cos \theta}{9+16 \sin 2 \theta}\right) d \theta$ is equal to :
   1. A.$3 \log _e 4$
   2. B.$4 \log _{\mathrm{e}} 3$
   3. C.$6 \log _e 4$
   4. D.$2 \log _e 3$
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   Answer: (B)

   $\begin{aligned} & I=\int_0^{\frac{\pi}{4}}\left(\frac{\sin \theta+\cos \theta}{9-16 \sin 2 \theta}\right) d \theta \\ & \text { Take } \sin \theta-\cos \theta=t \\ & (\cos \theta+\sin \theta) d \theta=d t \\ & (\sin \theta-\cos \theta)^2=t^2 \\ & \Rightarrow \sin 2 \theta=1-t^2 \\ & \theta=0 \rightarrow t=-1 \\ & \quad \theta=\frac{\pi}{4} \rightarrow t=0 \\ & I=\int_{-1}^0 \frac{d t}{9+16\left(1-t^2\right)} \\ & =\frac{1}{16} \int_{-1}^0 \frac{d t}{25} \frac{16}{16}-t^2 \\ & =\frac{1}{4}\left[\frac{1}{10} \log _{\left\lvert\, \frac{5+4 t}{}\right.}^{5-4 t}\right]_{-1}^0 \\ & =\frac{1}{40}\left[0+\log _e 9\right] \\ & I=\frac{\log _e 9}{40} \\ & 80 I=2 \log _e 9 \\ & 80 I=4 \log _e 3\end{aligned}$
3. Q3JEE Main 2024 (30 Jan, Shift 2)Composite Function
   Let $f:R\rightarrow R$ be a function defined $f\left(x\right)=\dfrac{x}{{\left(1+{x}^{4}\right)}^{1/4}}$ and $g\left(x\right)=f\left(f\left(f\left(f\left(x\right)\right)\right)\right)$ then $18{\int }_{0}^{\sqrt{2\sqrt{5}}}{x}^{2}g\left(x\right)dx$
   1. A.$33$
   2. B.$36$
   3. C.$42$
   4. D.$39$
   Show answer & solution

   Answer: (D)

   Given, $f\left(x\right)=\dfrac{x}{{\left(1+{x}^{4}\right)}^{1/4}}$ Now, finding $fof\left(x\right)=\dfrac{f\left(x\right)}{{\left(1+{f}^{4}\left(x\right)\right)}^{1/4}}=\dfrac{\dfrac{x}{{\left(1+{x}^{4}\right)}^{1/4}}}{{\left(1+\dfrac{{x}^{4}}{1+{x}^{4}}\right)}^{1/4}}=\dfrac{x}{{\left(1+2{x}^{4}\right)}^{1/4}}$ And $f\left(f\left(f\left(x\right)\right)\right)=\dfrac{f\left(x\right)}{{\left(1+2{f}^{4}\left(x\right)\right)}^{1/4}}=\dfrac{\dfrac{x}{{\left(1+{x}^{4}\right)}^{1/4}}}{{\left(1+\dfrac{2{x}^{4}}{1+{x}^{4}}\right)}^{1/4}}=\dfrac{x}{{\left(1+3{x}^{4}\right)}^{1/4}}$ $\Rightarrow f\left(f\left(f\left(f\left(x\right)\right)\right)\right)=\dfrac{x}{{\left(1+4{x}^{4}\right)}^{1/4}}=g\left(x\right)$ Now finding, $18{\int }_{0}^{\sqrt{2\sqrt{5}}}{x}^{2}g\left(x\right)dx$ $=18{\int }_{0}^{\sqrt{2\sqrt{5}}}\dfrac{{x}^{3}}{{\left(1+4{x}^{4}\right)}^{1/4}}dx$ Let $1+4{x}^{4}={t}^{4}$ $\Rightarrow 16{x}^{3}dx=4{t}^{3}dt$ So, the integral becomes $\dfrac{18}{4}{\int }_{1}^{3}\dfrac{{t}^{3}dt}{t}$ $=\dfrac{9}{2}{\int }_{1}^{3}{t}^{2}dt$ $=\dfrac{9}{2}{\left(\dfrac{{t}^{3}}{3}\right)}_{1}^{3}$ $=\dfrac{3}{2}\cdot 26=39$
4. Q4JEE Main 2023 (24 Jan, Shift 2)Basic Definite Integrals
   ${\int }_{\dfrac{3\sqrt{2}}{4}}^{\dfrac{3\sqrt{3}}{4}}\dfrac{48}{\sqrt{9-4{x}^{2}}}dx$ is equal to
   1. A.$\dfrac{\pi }{3}$
   2. B.$\dfrac{\pi }{2}$
   3. C.$\dfrac{\pi }{6}$
   4. D.$2\pi$
   Show answer & solution

   Answer: (D)

   Given integration is ${\int }_{\dfrac{3\sqrt{2}}{4}}^{\dfrac{3\sqrt{3}}{4}}\dfrac{48}{\sqrt{9-4{x}^{2}}}dx$ We know that $\int \dfrac{dx}{\sqrt{{a}^{2}-{x}^{2}}}={\sin }^{-1}\dfrac{x}{a}+C$ So, the equation can be re-written as ${\int }_{\dfrac{3\sqrt{2}}{4}}^{\dfrac{3\sqrt{3}}{4}}\dfrac{48}{2\sqrt{{\left(\dfrac{3}{2}\right)}^{2}-{x}^{2}}}dx=\dfrac{48}{2}\times {\left[{\sin }^{-1}\dfrac{2x}{3}\right]}_{\dfrac{3\sqrt{2}}{4}}^{\dfrac{3\sqrt{3}}{4}}$ $=24\times \left[{\sin }^{-1}\left(\dfrac{2}{3}\times \dfrac{3\sqrt{3}}{4}\right)-{\sin }^{-1}\left(\dfrac{2}{3}\times \dfrac{3\sqrt{2}}{4}\right)\right]$ $=24\times \left[{\sin }^{-1}\left(\dfrac{\sqrt{3}}{2}\right)-{\sin }^{-1}\left(\dfrac{1}{\sqrt{2}}\right)\right]$ $=24\times \left[\dfrac{\pi }{3}-\dfrac{\pi }{4}\right]=2\pi$
5. Q5JEE Main 2022 (28 Jul, Shift 1)Leibnitz Rule of Differentiation
   The minimum value of the twice differentiable function $f\left(x\right)={\int }_{0}^{x}{e}^{x-t}{f}^{'}\left(t\right)dt-\left({x}^{2}-x+1\right){e}^{x},x\in R$, is
   1. A.$-\dfrac{2}{\sqrt{e}}$
   2. B.$-2\sqrt{e}$
   3. C.$-\sqrt{e}$
   4. D.$\dfrac{2}{\sqrt{e}}$
   Show answer & solution

   Answer: (A)

   Given, $f\left(x\right)={\int }_{0}^{x}{e}^{x-t}{f}^{'}\left(t\right)dt-\left({x}^{2}-x+1\right){e}^{x}$ $\Rightarrow f\left(x\right)={e}^{x}{\int }_{0}^{x}{e}^{-t}{f}^{'}\left(t\right)dt-\left({x}^{2}-x+1\right){e}^{x}$ $\Rightarrow {e}^{-x}f\left(x\right)={\int }_{0}^{x}{e}^{-t}{f}^{'}\left(t\right)dt-\left({x}^{2}-x+1\right)$ Differentiate on both side w.r.t $x$ we get, ${e}^{-x}{f}^{'}\left(x\right)+\left(-f\left(x\right){e}^{-x}\right)={e}^{-x}{f}^{'}\left(x\right)-2x+1$ $\Rightarrow f\left(x\right)={e}^{x}\left(2x-1\right)$ $\Rightarrow {f}^{'}\left(x\right)={e}^{x}\left(2\right)+{e}^{x}\left(2x-1\right)$ $\Rightarrow {f}^{'}\left(x\right)={e}^{x}\left(2x+1\right)......\left(1\right)$ Now finding critical point we get, $2x+1=0\Rightarrow x=-\dfrac{1}{2}$ Now differentiating equation $\left(1\right)$ to check maxima and minima we get, ${f}^{"}\left(x\right)={e}^{x}\left(2\right)+\left(2x+1\right){e}^{x}$ $\Rightarrow {f}^{"}\left(x\right)={e}^{x}\left(2x+3\right)$ For $x=-\dfrac{1}{2},{f}^{"}\left(\dfrac{-1}{2}\right)\gt 0$, so it will give point of minima, Now minimum value will be given by, $f\left(\dfrac{-1}{2}\right)={e}^{-\dfrac{1}{2}}\left(-1-1\right)=-\dfrac{2}{\sqrt{e}}$