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$I=-\cot x \operatorname{cosec}^{3} x+\int \cot x\left(-3 \operatorname{cosec}^{2} x \cot x \operatorname{cosec} x\right) d x$

\section*{Topic I: Indefinite Integration}
1. If $\int \operatorname{cosec}^{5} x d x=\alpha \cot x \operatorname{cosec} x\left(\operatorname{cosec}^{2} x+\frac{3}{2}\right)+\beta \log _{e}\left|\tan \frac{x}{2}\right|+C$ where $\alpha, \beta \in R$ and $C$ is the constant of integration, then the value of $8(\alpha+\beta)$ equals $\qquad$ .
[JEE (Main) 4 April 2024 S II]
Sol.(1) $\int \operatorname{cosec}^{3} x \cdot \operatorname{cosec}^{2} x d x=1$
By applying integration by parts
$I=-\cot x \operatorname{cosec}^{3} x+\int \cot x\left(-3 \operatorname{cosec}^{2} x \cot x \operatorname{cosec} x\right) d x$
$I=-\cot x \operatorname{cosec}^{3} x-3 \int \operatorname{cosec}^{3} x\left(\operatorname{cosec}^{2} x-1\right) d x$
$I=-\cot x \operatorname{cosec}^{3} x-31+3 \int \operatorname{cosec}^{3} x d x$
$4 I=-\cot x \operatorname{cosec}^{3} x+3 \int \operatorname{cosec}^{3} x d x$ .
let $\mathrm{I}_{1}=\int \operatorname{cosec}^{3} \mathrm{x} d \mathrm{dx}=\int \operatorname{cosec} x \cdot \operatorname{cosec}^{2} \mathrm{x} \mathrm{dx}=-\operatorname{cosec} x \cot \mathrm{x}-\int \cot ^{2} \mathrm{x} \operatorname{cosec} \mathrm{xdx}$
$I_{1}=-\operatorname{cosec} x \cot x-\int\left(\operatorname{cosec}^{2} x-1\right) \operatorname{cosec} x d x=-\operatorname{cosec} x \cot x-\int \operatorname{cosec}^{3} x d x+\int \operatorname{cosec} x d x$
$\Rightarrow 21_{1}=-\operatorname{cosec} x \cot x+\ln \left|\tan \frac{x}{2}\right|$
$\Rightarrow I_{1}=-\frac{1}{2} \operatorname{cosec} x \cot x+\frac{1}{2} \ln \left|\tan \frac{x}{2}\right|$
From (1), $4 I=-\cot x \operatorname{cosec}^{3} x-\frac{3}{2} \operatorname{cosec} x \cot x+\frac{3}{2} \ln \left|\tan \frac{x}{2}\right|+4 c$
$I=-\frac{1}{4} \operatorname{cosec} x \cot x\left(\operatorname{cosec}^{2} x+\frac{3}{2}\right)+\frac{3}{8} \ln \left|\tan \frac{x}{2}\right|+c$
$\therefore \alpha=-\frac{1}{4}, \beta=\frac{3}{8} \Rightarrow 8(\alpha+\beta)=1$

$\begin{aligned} & I=-\cot x \operatorname{cosec} 3 x+\int \cot x\left(-3 \operatorname{cosec}^2 x \cot x \operatorname{cosec} x\right) d x \\ & I=-\cot x \operatorname{cosec}^3 x-3 \int \operatorname{cosec}^3 x\left(\operatorname{cosec}^2 x-1\right) d x \\ & I=-\cot x \operatorname{cosec}^3 x-3 I+3 \int \operatorname{cosec}^3 x d x \\ & A I=-\cot x \operatorname{cosec}^3 x+3 \int \operatorname{cosec}^3 x d x \ldots \ldots(1) \\ & \text { let } I_1=\int \operatorname{cosec} x d x=\int \operatorname{cosec} x \cdot \operatorname{cosec}^2 x d x=-\operatorname{cosec} x \cot x-\int \cot ^2 x \operatorname{cosec} x d x \\ & I_1=-\operatorname{cosec} x \cot x^3-\int\left(\operatorname{cosec}^2 x-1\right) \operatorname{cosec} x d x=-\operatorname{cosec} x \cot x-\int \operatorname{cosec}^3 x d x+\int \operatorname{cosec} x d x \\ & \Rightarrow 2 I_1=-\operatorname{cosec} x \cot x+\ln \left|\tan \frac{x}{2}\right| \\ & \Rightarrow I_1=-\frac{1}{2} \operatorname{cosec} x \cot x+\frac{1}{2} \ln \left|\tan \frac{x}{2}\right| \\ & \text { From }(1), 4 I=-\cot x \operatorname{cosec} 3 x-\frac{3}{2} \operatorname{cosec} x \cot x+\frac{3}{2} \ln \left|\tan \frac{x}{2}\right|+4 c \\ & I=-\frac{1}{4} \operatorname{cosec} x \cot x\left(\operatorname{cosec} 2 x+\frac{3}{2}\right)+\frac{3}{8} \ln \left|\tan \frac{x}{2}\right|+c \\ & \therefore \alpha=-\frac{1}{4}, \beta=\frac{3}{8} \Rightarrow 8(\alpha+\beta)=1\end{aligned}$

3. Let $\mathrm{I}(\mathrm{x})=\int \frac{6}{\sin ^{2} \mathrm{x}(1-\cot \mathrm{x})^{2}} d \mathrm{x}$ . If $\mathrm{I}(0)=3$ , then $\mathrm{I}\left(\frac{\pi}{12}\right)$ is equal to
(A) $2 \sqrt{3}$
(B) $6 \sqrt{3}$
(C) $\sqrt{3}$
(D) $3 \sqrt{3}$

Sol.(D) $I(x)=\int \frac{6 d x}{\sin ^{2} x(1-\cot x)^{2}}=\int \frac{6 \operatorname{cosec}^{2} x d x}{(1-\cot x)^{2}}$
Put $1-\cot \mathrm{x}=\mathrm{t} \Rightarrow \operatorname{cosec}^{2} \mathrm{xdx}=\mathrm{dt}$
$I=\int \frac{6 \mathrm{dt}}{\mathrm{t}^{2}}=-\frac{6}{\mathrm{t}}+\mathrm{c}$
$I(x)=-\frac{6}{1-\cot x}+c$
$\mathrm{I}(0)=-\frac{6}{\infty}+\mathrm{C}=3 \Rightarrow \mathrm{C}=3$

$\begin{aligned} & I(x)=3-\frac{6}{1-\cot x} \\ & I\left(\frac{\pi}{12}\right)=3-\frac{6}{1-(2+\sqrt{3})}=3+\frac{6}{\sqrt{3}+1}=3+\frac{6(\sqrt{3}-1)}{2}=3 \sqrt{3} \end{aligned}$

4. If $\int \frac{1}{\sqrt[5]{(x-1)^{4}(x+3)^{6}}} d x=A\left(\frac{\alpha x-1}{\beta x+3}\right)^{B}+C$ , where $C$ is the constant of integration, then the value of $\alpha+\beta+20 \mathrm{AB}$ is $\qquad$ –
[JEE (Main) 8 April 2024 S II]
Sol.(7) $\int \frac{1}{\sqrt[5]{(x-1)^{4}(x+3)^{6}}} d x=\int \frac{1}{\left(\frac{x-1}{x+3}\right)^{4 / 5}(x+3)^{2}} d x$
$\operatorname{Put}\left(\frac{x-1}{x+3}\right)=t \Rightarrow \frac{4}{(x+3)^{2}} d x=d t$
$\mathrm{I}=\frac{1}{4} \int \frac{1}{\mathrm{t}^{4 / 5}} \mathrm{dt}=\frac{1}{4} \frac{\mathrm{t}^{1 / 5}}{1 / 5}+\mathrm{C}=\frac{5}{4}\left(\frac{\mathrm{x}-1}{\mathrm{x}+3}\right)^{1 / 5}+\mathrm{C}$
$\mathrm{A}=\frac{5}{4}, \alpha=\beta=1, \mathrm{~B}=\frac{1}{5}$
$\alpha+\beta+20 \mathrm{AB}=2+20 \times \frac{5}{4} \times \frac{1}{5}=7$
5. Let $\int \frac{2-\tan x}{3+\tan x} d x=\frac{1}{2}\left(\alpha x+\log _{e}|\beta \sin x+\gamma \cos x|\right)+C$ , where $C$ is the constant of integration. Then $\alpha+\frac{\gamma}{\beta}$ is equal to
[JEE (Main) 9 April 2024 S I]
(A) 1
(B) 7
(C) 4
(D) 3

Sol.(C) $\int \frac{2-\tan x}{3+\tan x} d x=\int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x$
$2 \cos x-\sin x=A(3 \cos x+\sin x)+B(\cos x-3 \sin x)$
$\Rightarrow 2 \cos \mathrm{x}-\sin \mathrm{x}=\cos (3 \mathrm{~A}+\mathrm{B})+\sin \mathrm{x}(\mathrm{A}-3 \mathrm{~B})$
$3 \mathrm{~A}+\mathrm{B}=2$ and $\mathrm{A}-3 \mathrm{~B}=-1$
$\Rightarrow \mathrm{A}=\frac{1}{2}, \mathrm{~B}=\frac{1}{2}$
$\therefore \int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x=\int \frac{1}{2}+\frac{1}{2} \frac{\cos x-3 \sin x}{3 \cos x+\sin x}$
$=\frac{1}{2}(\mathrm{x}+\ln |3 \cos \mathrm{x}+\sin \mathrm{x}|)+\mathrm{C}$
we get $\alpha=1, \beta=1, \gamma=3$
$\therefore \alpha+\frac{\gamma}{\beta}=1+\frac{3}{1}=4$

6. The integral $\int \frac{\left(x^{8}-x^{2}\right) d x}{\left(x^{12}+3 x^{6}+1\right) \tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)}$ is equal to :
(A) $\log _{e}\left(\left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|\right)^{1 / 3}+C$
(B) $\log _{e}\left(\left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|\right)^{1 / 2}+C$
(C) $\log _{e}\left(\left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|\right)+C$
(D) $\log _{e}\left(\left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|\right)^{3}+C$

Sol.(A) Let $\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)=t \Rightarrow \frac{1}{1+\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} \cdot\left(3 x^{2}-\frac{3}{x^{4}}\right) d x=d t$
$\Rightarrow \frac{x^{6}}{x^{12}+3 x^{6}+1} \cdot \frac{3 x^{6}-3}{x^{4}} d x=d t \Rightarrow \frac{\left(x^{8}-x^{2}\right)}{\left(x^{12}+3 x^{6}+1\right)} d x=\frac{d t}{3}$
$I=\frac{1}{3} \int \frac{d t}{t}=\frac{1}{3} \ln |t|+C=\frac{1}{3} \log _{\mathrm{e}}\left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|+C$
$I=\log _{\mathrm{e}}\left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|^{1 / 3}+C$
Hence option (A) is correct
7. For $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , if $y(x)=\int \frac{\operatorname{cosec} x+\sin x}{\operatorname{cosec} x \sec x+\tan x \sin ^{2} x} d x$ and $\lim _{x \rightarrow \frac{\pi^{-}}{2}} y(x)=0$ then $y\left(\frac{\pi}{4}\right)$ is equal to

Sol.(C) $y(x)=\int \frac{\left(1+\sin ^{2} x\right) \cos x}{1+\sin ^{4} x} d x$ Put $\sin x=t, \cos x=d t$

$\begin{aligned} & y(x)=\int \frac{1+t^{2}}{t^{4}+1} d t=\int \frac{\frac{1}{t^{2}}+1}{\left(t-\frac{1}{t}\right)^{2}+2} d t=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{\left(t-\frac{1}{t}\right)}{\sqrt{2}}+C \\ & y(x)=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\sin x-\operatorname{cosec} x}{\sqrt{2}}\right)+C \end{aligned}$

$\lim _{\pi^{-}} y(x)=0=0+c \Rightarrow c=0$
$x \rightarrow \frac{\pi^{-}}{2}$
$y\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \tan ^{-1}\left(-\frac{1}{2}\right)$

NEW

Sol.(B) $I=\int \frac{\sin ^{3 / 2} x+\cos ^{3 / 2} x}{\sqrt{\sin ^{3} x \cos ^{3} x \sin (x-\theta)}} d x$
$I=\int \frac{\sin ^{3 / 2} x+\cos ^{3 / 2} x}{\sqrt{\sin ^{3} x \cos ^{3} x(\sin x \cos \theta-\cos x \sin \theta)}} d x$
$I=\int \frac{\sin ^{3 / 2} x}{\sin ^{3 / 2} x \cos ^{2} x \sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\cos ^{3 / 2} x}{\sin ^{2} x \cos ^{3 / 2} x \sqrt{\cos \theta-\cot x \sin \theta}} d x$
$I=\int \frac{\sec ^{2} x}{\sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\operatorname{cosec}^{2} x}{\sqrt{\cos \theta-\cot x \sin \theta}} d x$
let $\tan x \cos \theta-\sin \theta=t^{2} ; \sec ^{2} x d x=\frac{2 t d t}{\cos \theta}$
let $\cos \theta-\cot x \sin \theta=z^{2} ; \operatorname{cosec}^{2} x d x=\frac{2 z d z}{\sin \theta}$
$I=\int \frac{2 t d t}{t \cos \theta}+\int \frac{2 z d z}{z \sin \theta}=\frac{2 t}{\cos \theta}+\frac{2 z}{\sin \theta}$
$=2 \sec \theta \sqrt{\tan x \cos \theta-\sin \theta}+2 \operatorname{cosec} \theta \sqrt{\cos \theta-\cot x \sin \theta}$
Comparing we have $A=2 \sec \theta, B=2 \operatorname{cosec} \theta$ then $A B=8 \operatorname{cosec} 2 \theta$
9. Let $f(x)=\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x$ . If $f(3)=1 / 2\left(\log _{e} 5-\log _{e} 6\right)$ , then $f(4)$ is equal to
[JEE (Main) 25 Jan 2023 S I]
(A) $1 / 2\left(\log _{e} 19-\log _{e} 17\right)$ (B) $\log _{e} 17-\log _{e} 18$
(C) $1 / 2\left(\log _{e} 17-\log _{e} 19\right)(D) \log _{e} 19-\log _{e} 20$

Sol.(C) Put $\mathrm{x}^{2}=\mathrm{t}=2 \mathrm{xdx}=\mathrm{dt}$
$f(x)=\int \frac{d t}{(t+1)(t+3)}=\frac{1}{2} \int\left(\frac{1}{t+1}-\frac{1}{t+3}\right) d t$
$f(x)=\frac{1}{2}[\log (t+1)-\log (t+3)]+C$
$f(x)=\frac{1}{2}\left[\log \left(x^{2}+1\right)-\log \left(x^{2}+3\right)\right]+C$
put $x=3, f(3)=\frac{1}{2}(\ln 10-\ln 12)+C \Rightarrow C=0$
$f(4)=\frac{1}{2} \ln \left(\frac{17}{19}\right)$

Question bank 2013

EMI NEW

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