Question

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

• A. $\frac{1}{8}(x^2-1)+k$
• B. $\frac{1}{8}(x^2+1)+k$
• C. $\frac{x^2}{2}+k$
• D. $\frac{x}{2}+k$

Answer 1

Answer: (C)

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

Let $x=\cos 2\theta \cdot$
$dx=-2\sin 2\theta d\theta \cdot$
$-\int \cos \left\{2{\tan }^1\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}\right\}2\sin 2\theta \cdot d\theta$
$=-\int \cos \left\{2{\tan }^{-1}\right[\tan \theta \left]\right\}2\sin 2\theta \cdot d\theta$
$-\int \cos 2\theta \sin 2\theta .2d\theta \cdot$
$-2\sin 2\theta =d\cos 2\theta$
$\int \cos 2\theta \cdot \left(dcos2\theta \right)d\theta$
$=\frac{{\cos }^{2}2\theta }{2}+c$
$=\frac{x^2}{2}+c(\because \cos 2\theta =x)\cdot$