Question

Integrate the function $\frac{1}{\sqrt{{\sin }^{3}x\sin (x+\alpha )}}$ w.r.t $x$

Answer 1

Consider ${\sin }^{3}x\sin (x+\alpha )={\sin }^{3}x[\sin x\cos \alpha +\cos x\sin \alpha ]$ using $\sin (A+B)=\sin A\cos B+\cos A\sin B$

$={\sin }^{4}x\cos \alpha +\cos x{\sin }^{3}x\sin \alpha$

$={\sin }^{4}x\cos \alpha +\cos x{\sin }^{3}x\sin \alpha \times \frac{\sin x}{\sin x}$

$={\sin }^{4}x[\cos \alpha +\cos x\frac{\sin \alpha }{\sin x}]$

$={\sin }^{4}x[\cos \alpha +\cot x\sin \alpha ]$

Hence ${\sin }^{3}x\sin (x+\alpha )={\sin }^{4}x[\cos \alpha +\cot x\sin \alpha ]$ .......$\left(1\right)$

Now $\int \frac{dx}{\sqrt{{\sin }^{3}x\sin (x+\alpha )}}=\int \frac{dx}{\sqrt{{\sin }^{4}x[\cos \alpha +\cot x\sin \alpha ]}}$

$=\int \frac{dx}{\sqrt{{\sin }^{4}x}\sqrt{[\cos \alpha +\cot x\sin \alpha ]}}$

$=\int \frac{1}{{\sin }^{2}x}\times \frac{1}{\sqrt{\cos \alpha +\cot x.\sin \alpha }}dx$

Let $\cos \alpha +\cot x\sin \alpha =t$

$\Rightarrow dt=-{\csc }^{2}x\sin \alpha dx$

$\Rightarrow dx=\frac{{\sin }^{2}xdt}{-\sin \alpha }dt$

$\int \frac{1}{{\sin }^{2}x}\times \frac{1}{\sqrt{\cos \alpha +\cot x.\sin \alpha }}dx=\int \frac{1}{{\sin }^{2}x\sqrt{t}}\times \frac{1}{-\sin \alpha }\times {\sin }^{2}xdt$

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

$=\frac{-1}{\sin \alpha }(2\sqrt{t}+c)$

Put $t=\cos \alpha +\cot x\sin \alpha$

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