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Q.
$\int \frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}} d x, \alpha \neq n \pi, n \in Z$ is equal to
Integrals
Solution:
$\sin ^{3} x \sin (x+\alpha) =\sin ^{3} x(\sin x \cos \alpha+\cos x \sin \alpha)$
$=\sin ^{4} x(\cos \alpha+\cot x \sin \alpha)$
$I =\int \frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}} d x$
$=\int \frac{1}{\sin ^{2} x \sqrt{\cos \alpha+\cot x \sin \alpha}} d x$
$=\int \frac{\text{cosec}^{2} x}{\sqrt{\cos \alpha+\cot x \sin \alpha}} d x$
Putting $\cos \alpha+\cot x \sin \alpha=t$
and $-\text{cosec}^{2} x \sin \alpha d x=d t$, we have
$I =\int-\frac{1}{\sin \alpha \sqrt{t}} d t=-\frac{1}{\sin \alpha} \int t^{-1 / 2} d t$
$=-\frac{1}{\sin \alpha}\left(\frac{t^{1 / 2}}{1 / 2}\right)+C$
$\Rightarrow I =-2 \text{cosec}\, \alpha \sqrt{t}+C$
$=-2 \text{cosec}\, \alpha(\cos \alpha+\cot x \sin \alpha)^{1 / 2}+C$