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Integrate the function
$\int \frac{1}{\sqrt{\sin ^3 x \sin (x+a)}} d x=I(s a y)$
$\int \frac{1}{\sqrt{\sin ^3 x \sin (x+a)}} d x=I(s a y)$
Solution:
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Verified Answer
$\because \sin ^3 x \sin (x+a)=\sin ^3 x(\sin x \cos \alpha+\cos x \sin \alpha)$
$=\sin ^4 x(\cos \alpha+\cot x \sin \alpha)$
$I=\int \frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}} d x$
$=\int \frac{1}{\sin ^2 x \sqrt{\cos \alpha+\cot x \sin \alpha}} d x$
Put $\cos \alpha+\cot x \sin \alpha=t \Rightarrow-\operatorname{cosec}^2 x \sin \alpha=d t$
$I=\int-\frac{1}{\sin \alpha \sqrt{t}} d t=-\frac{1}{\sin \alpha} \int t^{-1 / 2} d t$
$=-\frac{1}{\sin \alpha}\left(\frac{t^{1 / 2}}{1 / 2}\right)+C=\frac{-2}{\sin \alpha} \cdot \sqrt{\frac{\sin (x+\alpha)}{\sin x}}+C$
$=\sin ^4 x(\cos \alpha+\cot x \sin \alpha)$
$I=\int \frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}} d x$
$=\int \frac{1}{\sin ^2 x \sqrt{\cos \alpha+\cot x \sin \alpha}} d x$
Put $\cos \alpha+\cot x \sin \alpha=t \Rightarrow-\operatorname{cosec}^2 x \sin \alpha=d t$
$I=\int-\frac{1}{\sin \alpha \sqrt{t}} d t=-\frac{1}{\sin \alpha} \int t^{-1 / 2} d t$
$=-\frac{1}{\sin \alpha}\left(\frac{t^{1 / 2}}{1 / 2}\right)+C=\frac{-2}{\sin \alpha} \cdot \sqrt{\frac{\sin (x+\alpha)}{\sin x}}+C$
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