JEE Main 2024 — 29 January, Shift 2 — Question 10

$\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^{3} x \cos ^{3} x \sin (x-\theta)}} d x$

$I=\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^{3} x \cos ^{3} x(\sin x \cos \theta-\cos x \sin \theta)}} d x$

$=\int \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x \cos ^{2} x \sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\cos ^{\frac{3}{2}} x}{\sin ^{2} x \cos ^{\frac{3}{2}} x \sqrt{\cos \theta-\cot x \sin \theta}} d x=$

$\int \frac{\sec ^{2} x}{\sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\operatorname{cosec}^{2} x}{\sqrt{\cos \theta-\cot x \sin \theta}} d x$ $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}$

Let For $\mathrm{I}_{1}$,

let $\tan \mathrm{x} \cos \theta-\sin \theta=\mathrm{t}^{2}$

$\sec ^{2} x d x=\frac{2 t d t}{\cos \theta}$ For $I_{2}$, let $\cos \theta-\cot x \sin \theta=z^{2}$

$$\operatorname{cosec}^{2} x d x=\frac{2 z d z}{\sin \theta}$$ $$\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}$$ $$\begin{aligned}& =\int \frac{2 \mathrm{tdt}}{\cos \theta \mathrm{t}}+\int \frac{2\mathrm{zdz}}{\sin \theta \mathrm{z}} & =\frac{2 \mathrm{t}}{\cos \theta}+\frac{2 \mathrm{z}}{\sin \theta}\end{aligned}$$

$I=2 \sec \theta \sqrt{\tan x \cos \theta-\sin \theta}+2 \operatorname{cosec} \theta \sqrt{\cos \theta-\cot x \sin \theta}$

$$\mathrm{AB}=8 \operatorname{cosec} 2 \theta$$