Question

Integrate w.r.to$x$

${\tan }^{-1}\left(\frac{\sqrt{1-\cos 2x}}{1+\cos 2x}\right)$

Answer 1

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

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

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

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

$=\int xdx=\frac{x^2}{2}.$

Hence, the answer is $\frac{x^2}{2}.$