Correct option is (4) \(8 \operatorname{cosec}(2 \theta)\)
\(\int \frac{\sin ^{\frac{3}{2}} \mathrm{x}+\cos ^{\frac{3}{2}} \mathrm{x}}{\sqrt{\sin ^{3} \mathrm{x} \cos ^{3} \mathrm{x} \sin (\mathrm{x}-\theta)}} \mathrm{dx}\)
\(I=\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^{3} x \cos ^{3} x(\sin x \cos \theta-\cos x \sin \theta)}} d x\)
\(=\int \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x \cos ^{2} x \sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\cos ^{\frac{3}{2}} x}{\sin ^{2} x \cos ^{\frac{3}{2}} x \sqrt{\cos \theta-\cot x \sin \theta}} d x\)
\(=\int \frac{\sec ^{2} x}{\sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\operatorname{cosec}^{2} x}{\sqrt{\cos \theta-\cot x \sin \theta}} d x\)
\(\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}\quad....{\{\text{Let}\}}\)
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 \(\cos \theta-\cot \mathrm{x} \sin \theta=\mathrm{z}^{2}\)
\(\operatorname{cosec}^{2} x d x=\frac{2 z d z}{\sin \theta}\)
\(\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}\)
\(=\int \frac{2 t d t}{\cos \theta t}+\int \frac{2 z d z}{\sin \theta z}\)
\(=\frac{2 \mathrm{t}}{\cos \theta}+\frac{2 \mathrm{z}}{\sin \theta}\)
\(=2 \sec \theta \sqrt{\tan x \cos \theta-\sin \theta}+2 \operatorname{cosec} \theta \sqrt{\cos \theta-\cot x \sin \theta}\)
Comparing
\(\mathrm{AB}=8 \operatorname{cosec} 2 \theta\)
