Question

Evaluate: ${\int }_0^1\frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx.$

Answer 1

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

Let $t={\sin }^{-1}x$

$\Rightarrow dt=\frac{dx}{\sqrt{1-x^2}}$

$\Rightarrow I=\int t\sin tdt$

$=t\int \sin tdt-\int \frac{dt}{dt}(\int \sin tdt)dt$

$=-t\cos t+\int \cos tdt$

$=-t\cos t+\sin t+c$

$\therefore I=-t\cos t+\sin t+c$

$I=-{\sin }^{-1}x$ $\cos$${\sin }^{-1}x$$+\sin {\sin }^{-1}x$ + c