JEE 2024 — Mathematics PYQ

$$\int \frac{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}{\sqrt{{\sin }^{3}x{\cos }^{3}x\sin (x-\theta )}}dx=A\sqrt{\cos \theta \tan x-\sin \theta }+B\sqrt{\cos \theta -\sin \theta \cot x}+C$$

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

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

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

$$\downarrow \downarrow$$
$$\begin{array}{rlrl}<br>\tan x& amp;=y& amp;-\cot x& amp;=z\\ <br>{\sec }^{2}xdx& amp;=dy& amp;{\text{cosec}}^{2}xdx& amp;=dz<br>\end{array}$$

$$I=\int \frac{dy}{\sqrt{y\cos \theta -\sin \theta }}+\int \frac{dz}{\sqrt{\cos \theta +z\sin \theta }}$$

$$=\frac{2\sqrt{y\cos \theta -\sin \theta }}{\cos \theta }+\frac{2\sqrt{\cos \theta +z\sin \theta }}{\sin \theta }$$

$$\therefore A=2\sec \theta$$
$$B=2\text{cosec}\theta$$

Calculating
$$AB=2\sec \theta \cdot 2\text{cosec}\theta$$
$$=\frac{4}{\sin \theta \cos \theta }=\frac{8}{2\sin \theta \cos \theta }$$
$$=\frac{8}{\sin 2\theta }=8\text{cosec}(2\theta )$$

$8\text{cosec}(2\theta )$