Question

${\int }_{-1}^1\frac{x{\sin }^{-1}x}{\sqrt{(1-x^2)}}dx=??$

Answer 1

Let $I={\int }_{-1}^1\frac{x{\sin }^{-1}x}{\sqrt{(1-x^2)}}dx$
And $f\left(x\right)=\frac{x{\sin }^{-1}x}{\sqrt{(1-x^2)}}$
$f(-x)=\frac{(-x){\sin }^{-1}(-x)}{\sqrt{[1-{(-x)}^2]}}=\frac{x{\sin }^{-1}x}{\sqrt{(1-x^2)}}=f\left(x\right)$
$I=2{\int }_0^1\frac{x{\sin }^{-1}x}{\sqrt{(1-x^2)}}dx$
Substitute $x=\sin \theta$
$\therefore I=2{\int }_0^{\pi /2}\theta \frac{\sin \theta \cos \theta d\theta }{\cos \theta }=2{\int }_0^{\pi /2}\theta \sin \theta d\theta =2\{\theta (-\cos \theta )+{\int }_0^1\cos \theta .1d\theta \}=2{\{-\theta \cos \theta +\sin \theta \}}_0^{\pi /2}=2[(0+1)-\left(0\right)]=2$