Question

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

Answer 1

$\int {\tan }^{-1}\left\{\sqrt{\left(\frac{1-\cos 2x}{1+\cos 2x}\right)}\right\}dx=\int {\tan }^{-1}\left\{\sqrt{\frac{2{\sin }^{2}x}{2{\cos }^{2}x}}\right\}dx(\because \cos 2x=1-2{\sin }^{2}xand\cos 2x=2{\cos }^{2}x-1)=\int {\tan }^{-1}\left(\sqrt{{\tan }^{2}x}\right)dx(\tan x=\frac{\sin x}{\cos x})=\int {\tan }^{-1}\left(\tan x\right)dx=\int xdx=\frac{x^2}{2}+C$