JEE Main 2024 — Indefinite Integration Question with Solution

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

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

$\Rightarrow I=\int \frac{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}{\sqrt{{\sin }^{3}x{\cos }^{4}x(\tan x\cos \theta -\sin \theta )}}dx$

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

$\Rightarrow I=\int \frac{1}{{\cos }^{2}x\sqrt{\tan x\cos \theta -\sin \theta }}dx+\int \frac{1}{{\sin }^{2}x\sqrt{\cos \theta -\cot x\sin \theta }}dx$

$\Rightarrow I=\int \frac{{\sec }^{2}x}{\sqrt{\tan x\cos \theta -\sin \theta }}dx+\int \frac{{cosec}^{2}x}{\sqrt{\cos \theta -\cot x\sin \theta }}dx$

So, $I=I_1+I_2 ....\left\{\mathrm{let}\right\}$

For $I_1$, let $\tan x\cos \theta -\sin \theta =t^2$

$\Rightarrow \cos \theta \times {\sec }^{2}xdx=2tdt$

$\Rightarrow {\sec }^{2}xdx=\frac{2tdt}{\cos \theta }$

For $I_2$, let $\cos \theta -\cot x\sin \theta =z^2$

$\Rightarrow {\mathrm{cosec}}^{2}x\times \sin \theta dx=2zdz$

$\Rightarrow {cosec}^{2}xdx=\frac{2zdz}{\sin \theta }$

$\Rightarrow I=\int \frac{2tdt}{\cos \theta \times t}+\int \frac{2zdz}{\sin \theta \times z}$

$\Rightarrow I=2\left[\left(\sec \theta \right)\int \left(1\right)dt+\mathrm{cosec}\theta \int \left(1\right)dz\right]$

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

$\Rightarrow A\times B=4\sec \theta \mathrm{cosec}\theta$

$\Rightarrow A\times B=\frac{2}{2}\times \frac{4}{\sin \theta \cos \theta }$

$\Rightarrow A\times B=8cosec2\theta$