$$ \begin{aligned} & \int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x \sin (x-\theta)}} d x \\ & I=\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x(\sin x \cos \theta-\cos x \sin \theta)}} d x \\ & =\int \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x \cos ^2 x \sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\cos ^{\frac{3}{2}} x}{\sin ^2 x \cos ^{\frac{3}{2}} x \sqrt{\cos \theta-\cot x \sin \theta}} d x= \\ & \quad \int \frac{\sec ^2 x}{\sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\operatorname{cosec}^2 x}{\sqrt{\cos \theta-\cot x \sin \theta}} d x \end{aligned} $$ 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 $\cos \theta-\cot \mathrm{x} \sin \theta=\mathrm{z}^2$ $$ \begin{aligned} & \operatorname{cosec}^2 x d x=\frac{2 z d z}{\sin \theta} \\ & I=I_1+I_2 \\ &=\int \frac{2 t d t}{\cos \theta t}+\int \frac{2 z d z}{\sin \theta z} \\ &=\frac{2 t}{\cos \theta}+\frac{2 z}{\sin \theta} \\ &=2 \sec \theta \sqrt{\tan x \cos \theta-\sin \theta}+2 \operatorname{cosec} \theta \sqrt{\cos \theta-\cot x \sin \theta} \end{aligned} $$ Comparing $$ \mathrm{AB}=8 \operatorname{cosec} 2 \theta $$