Question

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

• A. $2x^2+c$
• B. $\frac{x^2}{2}+c$
• C. $x$
• D. None of these

Answer 1

Answer: (B)

$\frac{x^2}{2}+c$

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

$\cos 2x=2{\cos }^{2}x-1=1-2{\sin }^{2}x$

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

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

$=\int xdx+C$

$=\frac{x^2}{2}+C$