QJEEM24Ma5849: JEE Main 2024 Mathematics - Class 12 JEEM24Ma5849 Solution: If , where is the integration constant, then is equal to

$\begin{aligned}& \text {} \int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x \sin (x-\theta)}} d x \\& I=\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x(\sin x \cos \theta-\cos x \sin \theta)}} d x \\& =\int \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x \cos ^2 x \sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\cos ^{\frac{3}{2}} x}{\sin ^2 x \cos ^{\frac{3}{2}} x \sqrt{\cos \theta-\cot x \sin \theta}} d x= \\& \int \frac{\sec ^2 x}{\sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\operatorname{cosec}^2 x}{\sqrt{\cos \theta-\cot x \sin \theta}} d x \\&\end{aligned}$

$\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2$ ...... {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 $\mathrm{I}_2$ , let $\cos \theta-\cot \mathrm{x} \sin \theta=\mathrm{z}^2$

$\operatorname{cosec}^2 x d x=\frac{2 z d z}{\sin \theta}$

$\begin{aligned}& I=I_1+I_2 \\& =\int \frac{2 t d t}{\cos \theta t}+\int \frac{2 z d z}{\sin \theta z} \\& =\frac{2 t}{\cos \theta}+\frac{2 z}{\sin \theta} \\& =2 \sec \theta \sqrt{\tan x \cos \theta-\sin \theta}+2 \operatorname{cosec} \theta \sqrt{\cos \theta-\cot x \sin \theta} \\& \text { Comparing } \\& \quad A B=8 \operatorname{cosec} 2 \theta\end{aligned}$