Integration JEE Main Questions: Important Questions, Previous Year Questions with Solutions

Integration JEE Questions with Solutions

The Integration JEE Main questions help you prepare effectively for the JEE Main exam. Solving integration JEE Main questions with solutions is one of the best ways to prepare for IIT JEE.

Question 1. Let e be the base of natural logarithm and let $f:\{1,2,3,4\} \rightarrow\left\{1, \mathrm{e}, \mathrm{e}^2, \mathrm{e}^3\right\}$ and
$g:\left\{1, e, e^2, e^3\right\} \rightarrow\left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right\}$ be two bijective functions such that $f$ is strictly decreasing and $\mathbf{g}$ is strictly increasing. If $\phi(x)=\left[f^{-1}\left\{g^{-1}\left(\frac{1}{2}\right)\right\}\right]^x$, then the area of the region $\mathrm{R}=\left\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^2 \leq \mathrm{y} \leq \phi(\mathrm{x})\right.$, $0 \leq x \leq 1\}$ is :

1. $\frac{3-\log _e(2)}{3 \log _e(2)}$

2. $\frac{1}{3 \log _e(2)}$

3. $3+\log _e(2)$

4. $\frac{3+\log _e(2)}{2+\log _e(3)}$

Correct Answer: Option (1)

Solution:

$\begin{aligned} & f(x)=e^{x-1} \\ & g(x)=\frac{1}{\log _e x+1} \\ & f^{-1}\left(g^{-1}(1 / 2)\right)=f^{-1}(e)=2 \\ & \phi(x)=2^x \\ & \therefore \int_0^1\left(2^x-x^2\right) d x=\left(\frac{2^x}{\ln 2}-\frac{x^3}{3}\right)_0^1=\frac{1}{\ln 2}-\frac{1}{3} \\ & =\frac{3-\ln 2}{3 \ln 2}\end{aligned}$

Question 2: The area of the region $\{(\mathrm{x}, \mathrm{y}): 0 \leq \mathrm{y} \leq 6-\mathrm{x}$, $\left.\mathrm{y}^2 \geq 4 \mathrm{x}-3, \mathrm{x} \geq 0\right\}$ is $:$

1. 8

2. 9

3. 12

4. 15

Correct Answer: Option (2)

Solution:

$A=\int_0^3\left(6-y-\frac{y^2+3}{4}\right) d y=9$

Question 3: The value of the integral $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{32 \cos ^4 x}{1+e^{\sin x}}\right) d x$ is :

1. $4 \pi+2$

2. $3 \pi+8$

3. $3 \pi+4$

4. $4 \pi+3$

Correct Answer: Option (2)

Solution:

Apply (P-5)

$I=\int_{-\pi / 4}^{\pi / 4} \frac{32 \cos ^4 \theta}{1+e^{-\sin \theta}} d \theta$

$\begin{aligned}
& \text { Add } 2 \mathrm{I}=\int_{-\pi / 4}^{\pi / 4} 32 \cos ^4 \theta \mathrm{~d} \theta=2 \int_0^{\pi / 4} 32 \cos ^4 \theta \mathrm{~d} \theta \\
& \mathrm{I}=32 \int_0^{\pi / 4} \cos ^4 \theta \mathrm{~d} \theta \\
& =8 \int_0^{\pi / 4}\left(2 \cos ^2 \theta\right)^2 \mathrm{~d} \theta=8 \int_0^{\pi / 4}(1+\cos 2 \theta)^2 \mathrm{~d} \theta \\
& =8 \int_0^{\pi / 4} 1+2 \cos 2 \theta+\cos ^2 2 \theta \mathrm{~d} \theta \\
& =8 \int_0^{\pi / 4} 1+2 \cos 2 \theta+\frac{1+\cos 4 \theta}{2} \mathrm{~d} \theta \\
& =8\left[\frac{3 \theta}{2}+\frac{2 \sin 2 \theta}{2}+\frac{\sin 4 \theta}{8}\right]_0^{\pi / 4} \\
& =8\left[\left(\frac{3 \pi}{8}+1+0\right)\right]=3 \pi+8
\end{aligned}$

Question 4: The value of the integral $\int_{-1}^1\left(\frac{x^3+|x|+1}{x^2+2|x|+1}\right) d x$ is equal to

1. $3 \log _e 2$

2. $2 \log _e 2$

3. $5 \log _e 3$

4. $3 \log _e 3$

Correct Answer: Option (2)

Solution: $\begin{aligned} & I=\int_{-1}^1 \frac{x^3+|x|+1}{x^2+2|x|+1} \cdot d x \\ & I=\int_{-1}^1 \frac{x^3}{x^2+2|x|+1} \cdot d x+\int_{-1}^1 \frac{|x|+1}{x^2+2|x|+1} \cdot d x \\ & I=0+\int_{-1}^1 \frac{|x|+1}{|x|^2+2|x|+1} \cdot d x \\ & 2 \int_0^1 \frac{1}{x+1} \cdot d x \\ & I=2[\ln |x+1|]_0^1 \\ & I=2 \ln (2)\end{aligned}$

Question 5: The area of the region $\left\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^2-8 \mathrm{x} \leq \mathrm{y} \leq-\mathrm{x}\right\}$ is:

1. $\frac{343}{6}$

2. $\frac{637}{6}$

3. $\frac{437}{6}$

4. $\frac{523}{6}$

Correct Answer: Option (1)

Solution: $\begin{aligned} & \text { Required area }=\int_0^7\left(7 x-x^2\right) d x \\ & =\left[\frac{7 x^2}{2}-\frac{x^3}{3}\right]_0^7=\frac{343}{6}\end{aligned}$

Question 6: If $\int_{\pi / 6}^{\pi / 4}\left(\cot \left(x-\frac{\pi}{3}\right) \cot \left(x+\frac{\pi}{3}\right)+1\right) d x=\alpha \log _e(\sqrt{3}-1)$, then $9 \alpha^2$ is equal to $\_\_\_\_$ .

1. 12

2. 13

3. 11

4. 14

Correct Answer: Option (1)

Solution:

$\begin{aligned} & \int_{\pi / 6}^{\pi / 4}\left(\left(\cot \left(\mathrm{x}-\frac{\pi}{3}\right) \cot \left(\mathrm{x}+\frac{\pi}{3}\right)+1\right) \mathrm{dx}\right. \\ & \text { using } \cot (\mathrm{A}-\mathrm{B})=\frac{\cot \mathrm{B} \cot \mathrm{A}+1}{\cot \mathrm{~B}-\cot \mathrm{A}} \\ & \Rightarrow \int_{\pi / 6}^{\pi / 4}-\frac{1}{\sqrt{3}}\left[\cot \left(\mathrm{x}-\frac{\pi}{3}\right)-\cot \left(\mathrm{x}+\frac{\pi}{3}\right)\right] \mathrm{dx} \\ & \Rightarrow \frac{-1}{\sqrt{3}}\left[\ln \left|\frac{\sin \left(\mathrm{x}-\frac{\pi}{3}\right)}{\sin \left(\mathrm{x}+\frac{\pi}{3}\right)}\right|\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}}\end{aligned}$

$\begin{aligned} & =\frac{-1}{\sqrt{3}}\left[\ln \left|\left(\frac{\sin 15^{\circ}}{\left.\sin 105^{\circ}\right)}\right)\right|-\ell \mathrm{n}\left|\frac{\sin 30^{\circ}}{\sin 90^{\circ}}\right|\right] \\ & =\frac{-1}{\sqrt{3}}\left[\ln \left(\tan 15^{\circ}\right)-\ell \mathrm{n}\left(\frac{1}{2}\right)\right. \\ & =\frac{-1}{\sqrt{3}}[\ln (2-\sqrt{3})+\ell \mathrm{n} 2] \\ & \Rightarrow \frac{-1}{\sqrt{3}} \ln (4-2 \sqrt{3})=\frac{-2}{\sqrt{3}} \ln (\sqrt{3}-1) \\ & \therefore \alpha=\frac{-2}{\sqrt{3}} \therefore 9 \alpha^2=12\end{aligned}$

Question 7: Let $f(x)=\left\{\begin{array}{lll}\frac{1}{3} & , & x \leq \pi / 2 \\ \frac{b(1-\sin x)}{(\pi-2 x)^2} & , & x>\pi / 2\end{array}\right.$, If $f$ is continuous at $\mathrm{x}=\pi / 2$, then the value of $\int_0^{3 \mathrm{~b}-6}\left|\mathrm{x}^2+2 \mathrm{x}-3\right| \mathrm{dx}$ is :

1. 5

2. 2

3. 3

4. 4

Correct Answer: Option (4)

Solution:

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{b\left(1-\cos \left(\frac{\pi}{2}-x\right)\right)}{4\left(\frac{\pi}{2}-x\right)^2}=\frac{1}{3} \Rightarrow b=\frac{8}{3}$

$\int_0^2\left|\mathrm{x}^2+2 \mathrm{x}-3\right| \mathrm{dx}=\int_0^2|(\mathrm{x}+3)(\mathrm{x}-1)| \mathrm{dx}$

put $\mathrm{x}-1=\mathrm{t}$

$\int_{-1}^1|(t+4) t| d t=\int_{-1}^0\left(-t^2-4 t\right) d t+\int_0^1\left(t^2+4 t\right) d t$

$\left(\frac{-t^3}{3}-2 t^2\right)_{-1}^0+\left(\frac{t^3}{3}+2 t^2\right)_0^1=4$

Question 8: The value of the integral
$\int_0^2 \frac{\sqrt{x\left(x^2+x+1\right)}}{(\sqrt{x+1})\left(\sqrt{x^4+x^2+1}\right)} d x$ is equal to :

1. $\frac{1}{3} \log _e(3-2 \sqrt{2})$

2. $\frac{2}{3} \log _{\mathrm{e}}(4+\sqrt{2})$

3. $\frac{2}{3} \log _e(3+2 \sqrt{2})$

4. $\frac{1}{3} \log _{\mathrm{e}}(1+6 \sqrt{2})$

Correct Answer: Option (3)

Solution: $\begin{aligned} & I=\int_0^2 \sqrt{\frac{x\left(x^2+x+1\right)}{(x+1)\left(x^2-x+1\right)\left(x^2+x+1\right)} d x} \\ & =\int_0^2 \sqrt{\frac{x}{x^3+1}} d x \\ & \left(x^{3 / 2}=t \Rightarrow \frac{3}{2} x^{1 / 2} d x=d t\right) \\ & =\frac{2}{3} \int_0^{2^{3 / 2}} \frac{d t}{\sqrt{t^2+1}} \\ & =\frac{2}{3}\left(\ln \left(t+\sqrt{t^2+1}\right)\right)_0^{2^{3 / 2}} \\ & =\frac{2}{3} \ln \left(2^{3 / 2}+3\right)\end{aligned}$

Question 9: The value of the integral $\int_0^{\infty} \frac{\log _{\mathrm{e}}(\mathrm{x})}{\mathrm{x}^2+4} \mathrm{dx}$ is:

1. $\frac{\pi \log _e(2)}{2}$

2. $\frac{\pi \log _e(2)}{4}$

3. $1+\pi \log _e(2)$

4. $2+\pi \log _{\mathrm{e}}(2)$

Correct Answer: Option (2)

Solution:

Put $x=2 t \Rightarrow d x=2 d t$

$\begin{aligned}
& \mathrm{I}=\int_0^{\infty} \frac{\ell \mathrm{n} 2 \mathrm{t}}{4 \mathrm{t}^2+4}(2 \mathrm{dt})=\frac{1}{2} \int_0^{\infty} \frac{\ell \mathrm{n} 2+\ell \mathrm{nt}}{\mathrm{t}^2+1} \mathrm{dt} \\
& =\frac{1}{2} \int_0^{\infty} \frac{\ell \mathrm{n} 2}{\mathrm{t}^2+1} \mathrm{dt}+\frac{1}{2} \int_0^{\infty} \frac{\ell \mathrm{nt}}{\mathrm{t}^2+1} \mathrm{dt} \\
& {\left[\frac{\ln 2}{2} \mathrm{tan}^{-1} \mathrm{t}\right]_0^{\infty}+\mathrm{I}_1} \\
& =\frac{\ell \mathrm{n} 2}{2} \cdot \frac{\pi}{2}+\mathrm{I}_1 \\
& \mathrm{I}_2=\frac{1}{2} \int_0^{\infty} \frac{\ell \mathrm{nt}}{\mathrm{t}^2+1} \mathrm{dt} \quad \mathrm{t}=\frac{1}{\mathrm{u}} \Rightarrow \mathrm{dt}=-\frac{\mathrm{du}}{\mathrm{u}^2} \\
& =\frac{1}{2} \int_{\infty}^0 \frac{\ell \mathrm{n}(1 / \mathrm{u})}{\frac{1}{\mathrm{u}^2}+1}\left(-\frac{\mathrm{du}}{\mathrm{u}^2}\right)
\end{aligned}$

Add $\Rightarrow \mathrm{I}_1=0$

$\mathrm{I}=\frac{\pi \ell \mathrm{n} 2}{4}$

Question 10: The integral $\int_0^1 \cot ^{-1}\left(1+\mathrm{x}+\mathrm{x}^2\right) \mathrm{dx}$ is equal to :

1. $2 \tan ^{-1} 2+\frac{1}{2} \log _e\left(\frac{5}{4}\right)+\frac{\pi}{2}$

2. $2 \tan ^{-1} 2+\frac{1}{2} \log _e\left(\frac{5}{4}\right)-\frac{\pi}{2}$

3. $2 \tan ^{-1} 2-\frac{1}{2} \log _e\left(\frac{5}{4}\right)+\frac{\pi}{2}$

4. $2 \tan ^{-1} 2-\frac{1}{2} \log _e\left(\frac{5}{4}\right)-\frac{\pi}{2}$

Correct Answer: Option (4)

Solution: $\begin{aligned}= & \int_0^1 \tan ^{-1}\left(\frac{1}{1+x+x^2}\right) d x \\ = & \int_0^1 \tan ^{-1}\left(\frac{(x+1)-x}{1+x(x+1)}\right) d x \\ = & \int_0^1\left(\tan ^{-1}(x+1)-\tan ^{-1} x\right) d x \\ = & \left(x \tan ^{-1}(x+1)-\frac{1}{2} \ell \mathrm{n}\left|1+(1+x)^2\right|+\tan ^{-1}(x+1)\right)_0^1 \\ & -\left(x \tan ^{-1} x-\frac{1}{2} \ell \mathrm{n}\left|1+x^2\right|\right)_0^1 \\ = & \left(2 \tan ^{-1} 2-\frac{\pi}{4}-\frac{1}{2} \ell \mathrm{n} \frac{5}{2}\right)-\left(\frac{\pi}{4}-\frac{1}{2} \ell \mathrm{n} 2\right) \\ = & 2 \tan ^{-1} 2-\frac{1}{2} \ell \mathrm{n}\left(\frac{5}{4}\right)-\frac{\pi}{2}\end{aligned}$

Question 11: Let $\int_{-2}^2(|\sin x|+[x \sin x]) d x=2(3-\cos 2)+\beta$, where [•] is the greatest integer function. Then $\beta \sin \left(\frac{\beta}{2}\right)$ equals.

1. 1

2. 2

3. 4

4. 8

Correct Answer: Option (2)

Solution: $\begin{aligned} & \int_{-2}^2(|\sin x|+[x \sin x]) d x \\ & \Rightarrow 2 \int_0^2(|\sin x|+[x \sin x]) d x \\ & \Rightarrow 2\left[\int_0^2 \sin x d x+\int_0^2[x \sin x] d x\right] \\ & 2\left([-\cos x]_0^2+\int_0^\alpha 0 d x+\int_\alpha^2 2 d x\right) \\ & 2(3-\cos 2-2-\alpha) \\ & 2(3-\cos 2)-2 \alpha \\ & \text { where } \alpha \text { is Root of } x \sin x=1 \\ & \text { Now } \operatorname{comparing} \beta=-2 \alpha \\ & \alpha=-\frac{\beta}{2} \\ & -\frac{\beta}{2} \sin \left(\frac{-\beta}{2}\right)=1 \\ & \beta \sin \left(\frac{\beta}{2}\right)=2\end{aligned}$

Question 12: The value of $\int_0^{20 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x$ is equal to:

1. $\frac{15 \pi}{2}$

2. $25 \pi$

3. $15 \pi$

4. $\frac{25 \pi}{2}$

Correct Answer: Option (3)

Solution: $\begin{aligned} & I=\int_0^{20 \pi}\left(1-2 \sin ^2 x \cos ^2 x\right) \cdot d x \\ & =20 \pi-\frac{1}{2} \int_0^{20 \pi} \sin ^2 2 x d x \\ & 20 \pi-\frac{1}{2}(40) \int_0^{\frac{\pi}{2}} \sin ^2 2 x \\ & 20 \pi-\frac{20}{2}\left(x-\frac{\sin 2 x}{2}\right)_0^{\frac{\pi}{2}} \\ & 20 \pi-10\left(\frac{\pi}{2}\right)=15 \pi\end{aligned}$

Question 13: Let $f(x)=\int \frac{d x}{x^{\left(\frac{2}{3}\right)}+2 x^{\left(\frac{1}{2}\right)}}$ be such that $f(0)=-26+24 \log _e(2)$. If $f(1)=a+b \log _e(3)$, where $a, b \in z$, then $a+b$ is equal to:

1. -18

2. -11

3. -5

4. -26

Correct Answer: Option (2)

Solution:

$f(x)=\int \frac{d x}{x^{2 / 3}+2 x^{1 / 2}}$

Put $x=t^6 \Rightarrow d x=6 t^5 d t$

$\begin{aligned}
& =\int \frac{6 t^5 d t}{t^4+2 t^3}=6 \int \frac{\left(t^2-4\right)+4}{t+2} d t \\
& =6\left[\int(t-2) d t+4 \int \frac{1}{t+2} d t\right] \\
& =6\left[\frac{\mathrm{t}^2}{2}-2 \mathrm{t}+4 \ln (\mathrm{t}+2)\right]+\mathrm{C} \\
& =3 \mathrm{x}^{1 / 3}-12 \mathrm{x}^{1 / 6}+24\left(\mathrm{n}\left(\mathrm{x}^{16}+2\right)+\mathrm{C}\right. \\
& \mathrm{f}(0)=24(\mathrm{n} 2+\mathrm{C}=-26+24(\mathrm{n} 2 \text { (given) } \\
& \Rightarrow \mathrm{C}=-26
\end{aligned}$

Now

$\begin{aligned}
& f(1)=-35+24(n 3=A+B(n 3 \text { (as given in ques.) } \\
& \Rightarrow A=-35 \& B=24 \\
& \Rightarrow A+B=-11
\end{aligned}$

Hence, the correct answer is option (2).

Question 14: The value of $\sum_{r=1}^{20}\left(\left|\sqrt{\pi\left(\int_0^r x|\sin \pi x| d x\right)}\right|\right)$ is $\_\_\_\_$

1. 210

2. 110

3. 112

4. 230

Correct Answer: Option (1)

Solution:

$\begin{aligned} & I_r=\int_0^r x|\sin \pi x| d x \\ & \text { King's Rule }=\int_0^r(r-x)|\sin \pi x| d x \\ & 2 I_r=\int_0^r r|\sin \pi x| d x \\ & I_r=\frac{r}{2} \int_0^r|\sin \pi x| d x \\ & I_1=\frac{1}{2} \int_0^1|\sin \pi x| d x=\frac{1}{2 \pi} \int_0^\pi|\sin t| d t=\frac{1}{2 \pi} \cdot 2 \\ & I_2=\frac{2}{2} \int_0^2|\sin \pi x| d x=\frac{2}{2 \pi} \int_0^{2 \pi}|\sin t| d t=\frac{2}{2 \pi} .4\end{aligned}$

$\begin{aligned} & \therefore S=\sqrt{\pi \cdot \frac{1}{2 \pi} \cdot 2}+\sqrt{\pi \cdot \frac{2}{2 \pi} \cdot 4}+\sqrt{\pi \cdot \frac{3}{2 \pi} \cdot 6 \ldots \ldots \ldots .+\sqrt{\pi \cdot \frac{20}{2 \pi} \cdot 2(20)}} \\ & =1+2+3+\ldots \ldots .20=\frac{20 \times 21}{2}=210\end{aligned}$

Hence, the correct answer is 210.

Question 15: If $\int\left(\frac{1-5 \cos ^2 x}{\sin ^5 x \cos ^2 x}\right) d x=f(x)+C$, where $C$ is the constant of integration, then $f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)$ is equal to

1. $\frac{1}{\sqrt{3}}(26-\sqrt{3})$

2. $\frac{2}{\sqrt{3}}(4+\sqrt{6})$

3. $\frac{4}{\sqrt{3}}(8-\sqrt{6})$

4. $\frac{1}{\sqrt{3}}(26+\sqrt{3})$

Correct Answer: Option (3)

Solution:

$\begin{aligned} & \int \frac{1-5 \cos ^2 x}{\sin ^2 x \cos ^2 x} d x \\ & \int \frac{\sec ^2 x d x}{\sin ^5 x}-\int \frac{5}{\sin ^5 x} d x \\ & \int \frac{1}{\sin ^5 x} \tan x-\int \frac{-5}{\sin ^6 x} \cos x \cdot \tan x d x-\int \frac{5}{\sin ^6 x} d x=\frac{\tan x}{\sin ^5 x}=f(x) \\ & f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)=\frac{32}{\sqrt{3}}-\frac{1}{\left(\frac{1}{\sqrt{2}}\right)^5}=\frac{32}{\sqrt{3}}-4 \sqrt{2}=\frac{4}{\sqrt{3}}(8-\sqrt{6})\end{aligned}$

Hence, the correct answer is option 3.

Question 16: If $\int\left(\frac{1-5 \cos ^2 x}{\sin ^5 x \cos ^2 x}\right) d x=f(x)+C$, where $C$ is the constant of integration, then $f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)$ is equal to

1. $\frac{1}{\sqrt{3}}(26-\sqrt{3})$

2. $\frac{2}{\sqrt{3}}(4+\sqrt{6})$

3. $\frac{4}{\sqrt{3}}(8-\sqrt{6})$

4. $\frac{1}{\sqrt{3}}(26+\sqrt{3})$

Correct Answer: Option (3)

Solution: $\begin{aligned} & \int \frac{1-5 \cos ^2 x}{\sin ^2 x \cos ^2 x} d x \\ & \int \frac{\sec ^2 x d x}{\sin ^5 x}-\int \frac{5}{\sin ^5 x} d x \\ & \int \frac{1}{\sin ^5 x} \tan x-\int \frac{-5}{\sin ^6 x} \cos x \cdot \tan x d x-\int \frac{5}{\sin ^6 x} d x=\frac{\tan x}{\sin ^5 x}=f(x) \\ & f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)=\frac{32}{\sqrt{3}}-\frac{1}{\left(\frac{1}{\sqrt{2}}\right)^5}=\frac{32}{\sqrt{3}}-4 \sqrt{2}=\frac{4}{\sqrt{3}}(8-\sqrt{6})\end{aligned}$

Hence, the correct answer is option 3.

Question 17: Let $f(t)=\int\left(\frac{\left(1-\sin \left(\log _e t\right)\right.}{1-\cos \left(\log _e t\right)}\right) d t, t>1$. if. $f\left(e^{\pi / 2}\right)=-e^{\pi / 2}$ and $f\left(e^{\pi / 4}\right)=\alpha e^{\pi / 4}$ then $\alpha$ equals:

1. $-1-2 \sqrt{2}$

2. $-1+\sqrt{2}$

3. $-1-\sqrt{2}$

4. $1+\sqrt{2}$

Correct Answer: Option (3)

Solution:

Let $\cos \mathrm{t}=\mathrm{x} \Rightarrow \mathrm{t}=\mathrm{e}^{\mathrm{x}} \Rightarrow \mathrm{dt}=\mathrm{e}^{\mathrm{x}} \mathrm{dx}$

$\begin{aligned} & G . I=\int \frac{1-\sin x}{2 \sin ^2 \frac{x}{2}} e^x d x=\frac{1}{2} \int\left(\cos e^2 \frac{x}{2}-2 \cot \frac{x}{2}\right) e^x d x=\frac{1}{2} e^x\left(-2 \cot \frac{x}{2}\right)+c \\ &\Rightarrow-t \cot \left(\frac{\operatorname{los} t}{2}\right)+c \\ & f\left(e^{-\pi / 2}\right)=-e^{\pi / 2} \quad \Rightarrow c=0 \\ & \therefore d\left(e^{\pi / 4}\right)=-e^{-\pi / 4}(\sqrt{2}+1) \\ & \therefore \alpha=-(\sqrt{2}+1)\end{aligned}$

Hence, the correct answer is option (3).

Question 18: Let $I(x)=\int \frac{3 d x}{(4 x+6)\left(\sqrt{4 x^2+8 x+3}\right)}$ and $I(0)=\frac{\sqrt{3}}{4}+20$.
If $I\left(\frac{1}{2}\right)=\frac{a \sqrt{2}}{b}+c$, where $a, b, c, \in N, \operatorname{gcd}(a, b)=1$, then $a+b+c$ is equal to:

1. 26

2. 23

3. 25

4. 24

Correct Answer: Option (1)

Solution:

$I(x)=\frac{3}{2} \int \frac{d x}{(2 x+3) \sqrt{(2 x+3)(2 x+1)}}=\frac{3}{2} \int \frac{d x}{(2 x+3)^{3 / 2}(2 x+1)^{1 / 2}}$

$\begin{aligned} & =\frac{3}{2} \int \frac{d x}{\left(\frac{2 x+3}{2 x+1}\right)^{3 / 2}(2 x+1)^2}\left(t=\frac{2 x+3}{2 x+1} \Rightarrow d t=\frac{-4}{(2 x+1)^2} d x\right) \\ & =\frac{3}{2} \int\left(\frac{2 x+3}{2 x+1}\right)^{-\frac{3}{2}} \frac{d x}{(2 x+1)^2}=\frac{-3}{8} \frac{\left(\frac{2 x+3}{2 x+1}\right)^{\frac{-3}{2}+1}}{\frac{-3}{2}+1}+C=\frac{3}{4}\left(\frac{2 x+1}{2 x+3}\right)^{\frac{1}{2}}+C \\ & I(0)=20+\frac{\sqrt{3}}{4} \Rightarrow c=20 \\ & I\left(\frac{1}{2}\right)=\frac{3 \sqrt{2}}{8}+20 \\ & a+b+c=3+8+20=26\end{aligned}$

Hence the correct answer is option (1).

Question 19: The value of the integral $\int_{\frac{\pi}{24}}^{\frac{5 \pi}{24}} \frac{\mathrm{~d} x}{1+\sqrt[3]{\tan 2 x}}$ is:

1. $\frac{\pi}{18}$

2. $\frac{\pi}{6}$

3. $\frac{\pi}{3}$

4. $\frac{\pi}{12}$

Correct Answer: Option (4)

Solution:

Let the given integral be $I$ :

$
I=\int_{\frac{\pi}{2 \pi}}^{\frac{5 \pi}{3 \pi}} \frac{d x}{1+\sqrt{\tan 2 x}}
$
The sum of the limits is $a+b=\frac{\pi}{24}+\frac{5 \pi}{24}=\frac{6 \pi}{24}=\frac{\pi}{4}$.
Using the property $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$, replace $x$ with $\left(\frac{\pi}{4}-x\right)$ :

$
\begin{aligned}
& I=\int_{\frac{\pi}{2 \pi}}^{\frac{\sin }{2 \pi}} \frac{d x}{1+\sqrt[3]{\tan 2\left(\frac{\pi}{4}-x\right)}} \\
& I=\int_{\frac{\pi}{2 \pi}}^{\frac{\sin }{2 \pi}} \frac{d x}{1+\sqrt[3]{\tan \left(\frac{\pi}{2}-2 x\right)}}
\end{aligned}
$
Since $\tan \left(\frac{\pi}{2}-\theta\right)=\cot \theta$ :

$
\begin{aligned}
& I=\int_{\frac{\pi}{2 \pi}}^{\frac{\frac{\pi}{2 \pi}}{2 \pi}} \frac{d x}{1+\sqrt[3]{\cot 2 x}} \\
& I=\int_{\frac{\pi}{2 \pi}}^{\frac{\pi}{2 \pi}} \frac{d x}{1+\frac{1}{\sqrt[3]{\tan 2 x}}} \\
& I=\int_{\frac{\pi}{2 \pi}}^{\frac{\pi}{2 \pi}} \frac{\sqrt[3]{\tan 2 x}}{1+\sqrt{\tan 2 x}} d x
\end{aligned}
$
Now, add the two expressions for $I$ :

$
\begin{aligned}
& 2 I=\int \frac{\frac{\pi}{\pi}}{\frac{\pi}{2 \pi}}\left(\frac{1}{1+\sqrt[3]{\tan 2 x}}+\frac{\sqrt[3]{\tan 2 x}}{1+\sqrt[3]{\tan 2 x}}\right) d x \\
& 2 I=\int \frac{\frac{\pi}{2}}{\frac{1+\sqrt[3]{\tan 2 x}}{1+\sqrt[3]{\tan 2 x}} d x} \\
& 2 I=\int \frac{\frac{\pi}{2 \pi}}{\frac{\pi}{24}} 1 d x \\
& 2 I=[x] \frac{\frac{\pi x}{\pi}}{24} \\
& 2 I=\frac{5 \pi}{24}-\frac{\pi}{24}=\frac{4 \pi}{24}=\frac{\pi}{6} \\
& I=\frac{\pi}{12}
\end{aligned}
$

Hence, the correct answer is option (4).

Question 20: The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{1}{[x]+4}\right) d x$, where [.] denotes the greatest integer function, is:

1. $\frac{1}{60}(\pi-7)$

2. $\frac{7}{60}(3 \pi-1)$

3. $\frac{7}{60}(\pi-3)$

4. $\frac{1}{60}(21 \pi-1)$

Correct Answer: Option (2)

Solution:

$I=\int_{-\frac{\pi } 2}^{-1} \frac{1}{-2+4} d x+\int_{-1}^0 \frac{1}{-1+4} d x+\int_0^1 \frac{1}{0+4} d x+\int_1^{\frac{\pi } 2} \frac{1}{1+4} d x$
$ I=\frac{1}{2}\left(-1+\frac{\pi}{2}\right)+\frac{1}{3}(0-(-1))+\frac{1}{4}(1-0)+\frac{1}{5}\left(\frac{\pi}{2}-1\right)$
$ I=\frac{\pi}{4}-\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{\pi}{10}-\frac{1}{5} $
$ I=\frac{7 \pi}{20}-\frac{7}{60}$
$I=\frac{7}{60}(3\pi-1)$

Hence, the correct answer is option (2).

Question 21: If $\int_0^1 4 \cot ^{-1}\left(1-2 x+4 x^2\right) d x=a \tan ^{-1}(2)-b \log _e(5)$, where $a \cdot b \in N$, then $(2 a+b)$ is equal to...

1. 9

2. 5

3. 6

4. 4

Correct Answer: Option (1)

Solution:

$\begin{aligned}
& 4 \int_0^1 \operatorname{Tan}^{-1}\left(\frac{1}{1-2 x+4 x^2}\right) d x \\
& 4 \int_0^1 \operatorname{Tan}^{-1}\left(\frac{2 x-(2 x-1)}{1+2 x+(2 x-1)}\right) d x \\
& I=4 \int_0^1 \operatorname{Tan}^{-1} 2 x . d x-4 \int_0^1 \operatorname{Tan}^{-1}(2 x-1) d x \ldots \ldots .(i) \\
& 4 \int_0^1 \operatorname{Tan}^{-1} 2 x . d x-4 \int_0^1 \operatorname{Tan}^{-1}(2(1-x)-1) d x \\
& I=4 \int_0^1 \operatorname{Tan}^{-1} 2 x . d x+4 \int_0^1 \operatorname{Tan}^{-1}(2 x-1) d x \ldots . .(i i) \\
& \text { (i) }+ \text { (ii) } 2 I=8 \int_0^1 \operatorname{Tan}^{-1} 2 x . d x
\end{aligned}$

By integration by parts $I=4 \operatorname{Tan}^1 2-\log 5$

$\therefore 2 a+b=8+1=9$

Hence, the correct answer is 9.

Question 22: $6 \int_0^\pi|(\sin 3 x+\sin 2 x+\sin x)| d x$ is equal to:

1. 17

2. 21

3. 24

4. 26

Correct Answer: Option (1)

Solution:

$\begin{aligned}
& \text { Sol: } 6 \int_0^\pi|\sin 3 x+\sin 2 x+\sin x| d x \\
& =6 \int_0^\pi|\sin 2 x+2 \sin 2 x \cos x| d x=6 \int_0^\pi|\sin 2 x(1+2 \cos x)| d x \\
& =6 \int_0^\pi|2 \cos x(1+2 \cos x)| \sin x d x \quad \cos x=t \\
& =12 \int_1^{-1}|t(1+2 t)|-d t \quad-\sin x \cdot d x=d t
\end{aligned}$

$\begin{aligned}
& =12 \int_{-1}^1|t(1+2 t)| d t=12\left[\int_{-1}^{-1 / 2}\left(t+2 t^2\right) d t-\int_{-1 / 2}^0\left(t+2 t^2\right) d t+\int_0^1\left(t+2 t^2\right) d t\right] \\
= & 12\left[\left(\frac{t^2}{2}+\frac{2 t^3}{3}\right)_{-1}^{-1 / 2}-\left(\frac{t^2}{2}+\frac{2 t^3}{3}\right)_{-1 / 2}+\left(\frac{t^2}{2}+\frac{2 t^3}{3}\right)\right] \\
= & 12\left[\left(\frac{1}{8}-\frac{1}{12}\right)-\left(\frac{1}{2}-\frac{2}{3}\right)+\left(\frac{1}{8}-\frac{1}{12}\right)+\left(\frac{1}{2}+\frac{2}{3}\right)\right] \\
= & 12\left[\frac{2}{8}-\frac{2}{12}+\frac{4}{3}\right]=12\left[\frac{1}{4}-\frac{1}{6}+\frac{4}{3}\right]=3-2+16=17
\end{aligned}$

OR,

$\begin{aligned}
& 6 \int_0^\pi|2 \sin 2 x \cos x+\sin 2 x| d x \\
& 6 \int_0^\pi\left|4 \sin x \cos ^2 x+2 \sin \cos x\right| d x
\end{aligned}$

$I=12 \int_0^\pi \sin x\left|2 \cos ^2 x+\cos x\right| d x \quad \text { Put } \cos x=t,-\sin x d x=d t$

Hence, the correct answer is 17.

Question 23: If a function $f(x)$ satisfies a condition $f(x)=\cos x+\int_0^{\pi / 2} \sin x \cdot \sin y f(y) dy$, then $f\left(\frac{\pi}{4}\right)$ is equal to:

1. $\frac{1}{3 \sqrt{2}}\left(\frac{2+\pi}{2-\pi}\right)$

2. $\frac{4-\pi}{3 \sqrt{2}}$

3. $\frac{2-\pi}{2 \sqrt{2}}$

4. $\frac{1}{\sqrt{2}}\left(\frac{6-\pi}{4-\pi}\right)$

Correct Answer: Option (4)

Solution:

Let, $f(x)=\cos x+k \sin x$
where $k=\int_0^{\pi / 2} \sin y \cdot f(y) d y$

$
\begin{aligned}
& \Rightarrow k=\int_0^{\pi / 2}\left(\sin y \cdot \cos y+k \sin ^2 y\right) d y=-\frac{\cos 2 y}{4}+\left.\frac{k}{2}\left(y-\frac{\sin 2 y}{2}\right)\right|_0 ^{\pi / 2} \\
& \Rightarrow k=\frac{1}{2}+\frac{k \pi}{4} \Rightarrow k=\frac{2}{4-\pi}
\end{aligned}
$

Hence $f(x)=\cos x+\left(\frac{2}{4-\pi}\right) \sin x$

Question 24: Evaluate $\int \frac{\sec x}{2 \tan x+\sec x-1} d x$

1. $-\frac{1}{2} \ln \left|1+2 \tan \frac{x}{2}\right|+c$

2. $\frac{1}{2} \ln \left|1+3 \tan \frac{x}{2}\right|+c$

3. $-\frac{1}{2} \ln \left|1+2 \cot \frac{x}{2}\right|+c$

4. $\frac{1}{2} \ln \left|1+3 \cot \frac{x}{2}\right|+c$

Correct Answer: Option (3)

Solution:

$\begin{aligned}
& I=\int \frac{\sec x d x}{2 \tan x+\sec x-1} \\
& =\int \frac{d x}{2 \sin x+1-\cos x}=\int \frac{d x}{4 \sin \frac{x}{2} \cos \frac{x}{2}+2 \sin ^2 \frac{x}{2}} \\
& =\frac{1}{2} \int \frac{\operatorname{cosec}^2 \frac{x}{2} d x}{2 \cot \frac{x}{2}+1}
\end{aligned}$

Let $2 \cot \frac{\mathrm{x}}{2}=\mathrm{t} \Rightarrow-\operatorname{cosec}^2 \frac{\mathrm{x}}{2} \mathrm{dx}=\mathrm{dt}$
Hence $I=-\frac{1}{2} \int \frac{d t}{t+1}=-\frac{1}{2} \ln |1+t|+$ constant

$=-\frac{1}{2} \ln \left|1+2 \cot \frac{x}{2}\right|+\text { c }$

Hence, the correct answer is option (3).

Question 25: $4 \int_{0}^{1}\left(\frac{1}{\sqrt{3+\mathrm{x}^{2}}+\sqrt{1+\mathrm{x}^{2}}}\right) \mathrm{dx}-3 \log _{\mathrm{e}}(\sqrt{3})$ is equal to:

1. $2+\sqrt{2}+\log _{\mathrm{e}}(1+\sqrt{2})$

2. $2-\sqrt{2}-\log _{\mathrm{e}}(1+\sqrt{2})$

3. $2+\sqrt{2}-\log _{\mathrm{e}}(1+\sqrt{2})$

4. $2-\sqrt{2}+\log _{\mathrm{e}}(1+\sqrt{2})$

Correct Answer: Option (2)

Solution:

Given the expression:

$4 \int_0^1 \frac{1}{\sqrt{3 + x^2} + \sqrt{1 + x^2}} \, dx - 3 \log_e(\sqrt{3})$

First, rationalize the denominator inside the integral by multiplying numerator and denominator by

$\sqrt{3 + x^2} - \sqrt{1 + x^2}$

This gives:

$\int_0^1 \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{(\sqrt{3 + x^2})^2 - (\sqrt{1 + x^2})^2} \, dx$

Since

$(\sqrt{3 + x^2})^2 - (\sqrt{1 + x^2})^2 = (3 + x^2) - (1 + x^2) = 2$

the integral becomes

$\int_0^1 \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{2} \, dx = \frac{1}{2} \int_0^1 \left(\sqrt{3 + x^2} - \sqrt{1 + x^2}\right) dx$

Therefore, the original expression is

$4 \times \frac{1}{2} \int_0^1 \left(\sqrt{3 + x^2} - \sqrt{1 + x^2}\right) dx - 3 \log_e(\sqrt{3}) = 2 \int_0^1 \left(\sqrt{3 + x^2} - \sqrt{1 + x^2}\right) dx - \frac{3}{2} \log_e 3$

Split the integral:

$2 \left[ \int_0^1 \sqrt{3 + x^2} \, dx - \int_0^1 \sqrt{1 + x^2} \, dx \right] - \frac{3}{2} \log_e 3$

Recall the integral formula

$\int \sqrt{x^2 + c} \, dx = \frac{x}{2} \sqrt{x^2 + c} + \frac{c}{2} \ln \left( x + \sqrt{x^2 + c} \right) + C$

Apply for $c=3$:

$\int_0^1 \sqrt{3 + x^2} \, dx = \left[ \frac{x}{2} \sqrt{x^2 + 3} + \frac{3}{2} \ln \left( x + \sqrt{x^2 + 3} \right) \right]_0^1$

Evaluate at limits:

$= \frac{1}{2} \sqrt{4} + \frac{3}{2} \ln (1 + 2) - \left( 0 + \frac{3}{2} \ln \sqrt{3} \right) = 1 + \frac{3}{2} \ln 3 - \frac{3}{4} \ln 3 = 1 + \frac{3}{4} \ln 3$

Similarly for $c=1$:

$\int_0^1 \sqrt{1 + x^2} \, dx = \left[ \frac{x}{2} \sqrt{x^2 + 1} + \frac{1}{2} \ln \left( x + \sqrt{x^2 + 1} \right) \right]_0^1$

Evaluate at limits:

$= \frac{1}{2} \sqrt{2} + \frac{1}{2} \ln (1 + \sqrt{2}) - 0 = \frac{\sqrt{2}}{2} + \frac{1}{2} \ln (1 + \sqrt{2})$

Substitute back:

$2 \left( 1 + \frac{3}{4} \ln 3 - \frac{\sqrt{2}}{2} - \frac{1}{2} \ln (1 + \sqrt{2}) \right) - \frac{3}{2} \ln 3$

Simplify:

$= 2 - \sqrt{2} + \frac{3}{2} \ln 3 - \ln (1 + \sqrt{2}) - \frac{3}{2} \ln 3 = 2 - \sqrt{2} - \ln (1 + \sqrt{2})$

Hence, the value of the original expression is

$2 - \sqrt{2} - \ln (1 + \sqrt{2})=2-\sqrt{2}-\log _{\mathrm{e}}(1+\sqrt{2})$

Hence, the correct answer is option (2).

Integration JEE Last 10 Years' Most Asked Topics

JEE Main Integration questions over the last 10 years have covered various topics. The table below shows the most asked JEE Main and JEE Advanced integration question topics.

Also Check: JEE Main 2027 Important Formulas

Important Integration Formulas for JEE Main

The important formulas to solve Integration questions in JEE Main are given below. Students should learn and revise these formulas regularly.

1. Power Rule of Integration
   $\int x^n,dx = \frac{x^{n+1}}{n+1}+C,\quad n\neq -1$

2. Logarithmic Integration Formula
   $\int \frac{1}{x},dx = \ln|x|+C$

3. Exponential Function Formula
   $\int e^x,dx = e^x+C$

4. General Exponential Formula
   $\int a^x,dx = \frac{a^x}{\ln a}+C,\quad a>0,\ a\neq1$

5. Integration of Sine Function
   $\int \sin x,dx = -\cos x+C$

6. Integration of Cosine Function
   $\int \cos x,dx = \sin x+C$

7. Integration of Secant Square
   $\int \sec^2 x,dx = \tan x+C$

8. Standard Inverse Tangent Formula
   $\int \frac{1}{1+x^2},dx = \tan^{-1}x+C$

9. Standard Inverse Sine Formula
   $\int \frac{1}{\sqrt{1-x^2}},dx = \sin^{-1}x+C$

10. Integration by Parts Formula
    $\int u,dv = uv-\int v,du$

Best Books for Integration JEE Main Questions

While preparing for IIT JEE, it is very important to follow the best books because they help properly cover all the concepts. Refer to the books given below for solving Integration JEE Main questions and answers.

Related JEE Main Maths Preparation Resources