Question

Evalaute the integral
${\int }_0^1{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx$

• A. $\frac{\pi }{4}-\log 2$
• B. $\frac{\pi }{2}+\log 2$
• C. $\frac{\pi }{2}-\log 2$
• D. $\frac{\pi }{4}+\log 2$

Answer 1

Answer: (C)

$\frac{\pi }{2}-\log 2$

${\int }_0^1{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx$
$={\left[x{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\right]}_0^1-{\int }_0^1\frac{x2(x^2-1)}{(x^4-1)}dx$
$=\frac{\pi }{2}-0-2{\int }_0^1\frac{x}{x^2+1}dx$
$=\frac{\pi }{2}-{\left[\log (x^2+1)\right]}_0^1=\frac{\pi }{2}-\log 2$