Find $\int \frac{dx}{\sqrt{\sin^3 x \cos(x - \alpha)}}$.

$-2 \frac{\sqrt{\cot x \cos \alpha + \sin \alpha}}{\sqrt{\cos \alpha}} + c$

The correct answer is Option (1) → $-2 \frac{\sqrt{\cot x \cos \alpha + \sin \alpha}}{\sqrt{\cos \alpha}} + c$

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

$= \int \frac{dx}{\sin^2 x \sqrt{\frac{\cos x}{\sin x} \cos \alpha + \sin \alpha}}$

$= \int \frac{\text{cosec}^2 x \, dx}{\sqrt{\cot x \cos \alpha + \sin \alpha}}$

$=\frac{1}{\sqrt{\cos \alpha}}\int \frac{\text{cosec}^2 x \, dx}{\cot x+\tan \alpha}$

Let $\cot x +\tan \alpha=t$

$-\text{cosec}^2 x dx=dt$

$= \frac{1}{\sqrt{\cos \alpha}} \int \frac{-dt}{\sqrt{t}}$

$= -\frac{2\sqrt{t}}{\sqrt{\cos \alpha}} + c$

$=-2 \frac{\sqrt{\cot x \cos \alpha + \sin \alpha}}{\sqrt{\cos \alpha}} + c$