Question

If $I={\int }_{-1}^1\frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx$ then $I$ equals

• A. $1$
• B. $2$
• C. $\pi$
• D. $2\pi$

Answer 1

The correct option is B $2$

$I={\int }_{-1}^1\frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx$
Substitute $x=\sin t\Rightarrow dx=\cos tdt$
$\therefore I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{t\sin t}{\cos t}\cos tdt$
$\Rightarrow I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}t\sin tdt={[-t\cos t]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}+{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos tdt$
$=0+{\left[\sin t\right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}=1+1=2$