Question

$\int {\tan }^{-1}\left\{\sqrt{\left(\frac{1-\cos 2x}{1+\cos 2x}\right)}\right\}dx,0<x<\frac{\pi }{2}$ is equal to

• A. $\frac{x^2}{2}+C$
• B. $\frac{x^3}{3}+C$
• C. $\frac{-x^2}{2}+C$
• D. $\frac{x^4}{4}+C$

Answer 1

The correct option is A $\frac{x^2}{2}+C$

$\int {\tan }^{-1}\left\{\sqrt{\left(\frac{1-\cos 2x}{1+\cos 2x}\right)}\right\}dx=\int {\tan }^{-1}\left\{\sqrt{\left(\frac{2{\sin }^{2}x}{2{\cos }^{2}x}\right)}\right\}dx$
$\{\because 1-\cos 2x=2{\sin }^{2}x,1+\cos 2x=2{\cos }^{2}x\}$
$\Rightarrow \int {\tan }^{-1}\left(|\tan x|\right)dx$
$\{\because x$ is in first quadrant ,hence $|\tan x|=\tan x\}$
$\Rightarrow \int {\tan }^{-1}\left(\tan x\right)dx=\int xdx=\frac{x^2}{2}+C$