Question

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

Answer 1

$I=\int {\tan }^{-1}\sqrt{\frac{1-\sin x}{1+\sin x}}d{x}_n$

$=\int {\tan }^{-1}\sqrt{\frac{{\cos }^{2}x/2+{\sin }^{2}x/2-2\cos x/2\sin x/2}{{\cos }^{2}x/2+{\sin }^{2}x/2+2\cos x/2\sin x/2}}dx$

$=\int {\tan }^{-1}\sqrt{\frac{(\cos x/2+\sin x/2{)}^2}{(\cos x/2+\sin x{)}^2}}dx$

$=\int {\tan }^{-1}\left(\frac{\cos x/2-\sin x/2}{\cos x/2+\sin x/2}\right)dx$

$=\int {\tan }^{-1}\left(\tan (\frac{\pi }{4}-\frac{x}{2})\right)dx$

$=\int \frac{\pi }{4}-\frac{x}{2}dx$

$=[\frac{\pi }{4}x-\frac{x^2}{4}]$.