Question

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

• A. $\frac{1}{8}(x^2-1)+k$
• B. $\frac{1}{2}{x}^2+k$
• C. $\frac{1}{2}x+k$
• D. none of these

Answer 1

The correct option is B $\frac{1}{2}{x}^2+k$

$\int \cos \left\{2{\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right\}dx$
Put $x=\cos \theta$
$\Rightarrow dx=-\sin \theta d\theta$
$=-\int \cos \left\{2{\tan }^{-1}\sqrt{{\tan }^2\left(\frac{\theta }{2}\right)}\right\}\sin \theta d\theta$
$=-\int \cos \theta \sin \theta d\theta$
$=\int xdx$
$=\frac{x^2}{2}+k$