Question

$\int \frac{1}{\sqrt{si{n}^{3}xsin(x+\alpha )}}dx.$

Answer 1

$LetI=\int \frac{1}{\sqrt{si{n}^{3}xsin(x+\alpha )}}dx=\int \frac{1}{\sqrt{si{n}^{3}x(sinxcos\alpha +cosxsin\alpha )}}dx[\because sin(A+B)=sinAcosB+cosAsinB]=\int \frac{1}{\sqrt{si{n}^{4}x(cos\alpha +cotxsin\alpha )}}dx=\int \frac{1}{si{n}^{2}x\sqrt{cos\alpha +cotxsin\alpha }}dx=\int \frac{cose{c}^{2}x}{\sqrt{cos\alpha +cotxsin\alpha }}dxPutcos\alpha +cotxsin\alpha =t\Rightarrow -cose{c}^{2}xsin\alpha dx=dt\Rightarrow dx=\frac{dt}{-cose{c}^{2}xsin\alpha }\therefore I=\frac{-1}{sin\alpha }\int \frac{dt}{\sqrt{t}}=\frac{-1}{sin\alpha }\left[2\sqrt{t}\right]+C=\frac{-1}{sin\alpha }\left[2\sqrt{cos\alpha +cotxsin\alpha }\right]+C=\frac{-2}{sin\alpha }\sqrt{\cos \alpha +\frac{cosxsin\alpha }{sinx}}+C=\frac{-2}{sin\alpha }\sqrt{\frac{sinxcos\alpha +cosxsin\alpha }{sinx}}+C=-\frac{2}{sin\alpha }\sqrt{\frac{sin(x+\alpha )}{sinx}}+C(\because sin(A+B)=sinAcosB+cosAsinB)$