Question

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

• A. $\frac{1}{8}(x^2-1)+c$
• B. $\frac{x^4}{4}+c$
• C. $\frac{x}{2}+c$
• D. $\frac{x}{4}+c$
• E. $\frac{x^2}{2}+c$

Answer 1

Answer: (E)

Let $I=\int \cos \left\{2{\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right\}dx$ Put $x=cos\theta$
$\therefore$ $I=\int \cos \left\{2{\tan }^{-1}\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\right\}dx$
$=\int \cos \left\{2{\tan }^{-1}\left(\tan \frac{\theta }{2}\right)\right\}dx$
$=\int \cos \theta dx=\int xdx=\frac{x^2}{2}+c$