Question

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

• A. $2x^2+6$
• B. $x^2+6$
• C. $\frac{x^2}{2}+6$
• D. $2x+6$

Answer 1

Answer: (C)

$\frac{x^2}{2}+6$

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

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

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

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

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

$\int xdx=\frac{x^2}{2}+6$ (6 is also a constant)