Question

$\int {\tan }^{-1}\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}dx$, where $0<x<\frac{\pi }{2}$. is equal to

Answer 1

Consider given the integration,

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

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

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

$=\int {\tan }^{-1}\tan xdx$

$=\int 1dx$

$I=x+C$

Hence, this is the answer,