Question

If $\int \frac{dx}{\sqrt{{\sin }^{3}x\sin (x+\alpha )}}=-k\text{cosec}\alpha {\text{cosec}}^{2}x\sqrt{{\sin }^{3}x\sin (\alpha +x)}$, then find the value of $k$.

Answer 1

$\int \frac{dx}{\sqrt{{\sin }^{3}x.\sin (x+a)}}=\int \frac{dx}{\sqrt{{\sin }^{4}x.\frac{\sin (x+a)}{\sin x}}}$

$=\int \frac{{\csc }^{2}xdx}{\sqrt{\frac{\sin (x+a)}{\sin x}}}$

$=\int \frac{{\csc }^{2}xdx}{\sqrt{\cos a+\cot x\sin a}}$

Let $\cot x=t$

$-{\csc }^{2}xdx=dt$

$\Rightarrow {\csc }^{2}xdx=-dt$

$=\int \frac{-dt}{\sqrt{\cos a+t\sin a}}$

$=\frac{-2\sqrt{\cos a+t\sin a}}{\sin a}as\int \frac{1}{\sqrt{ax}}=\frac{2\sqrt{ax}}{a}$

$=-2\csc a\sqrt{\cos a+\cot x\sin a}$

$=-2\csc a\sqrt{\frac{\sin (x+a)}{\sin x}}$

$=-2\csc a\sqrt{\frac{\sin (x+a)}{{\sin }^{4}x}{\sin }^{3}x}$

$=-2\csc a{\csc }^{2}x\sqrt{{\sin }^{3}x\sin (x+a)}$

So, $k=2$