Question

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

Answer 1

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

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

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

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

Substituting this value,

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

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

Hence, we take

First function : $f\left(x\right)=t$

Second function : $g\left(x\right)=\sin t$

$I=t\int \sin tdt-\int \frac{d\left(t\right)}{dt}\int \sin tdtdt$

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

$I=-t\cos t+\sin t+C$

$I=x-\sqrt{1-x^2}{\sin }^{-1}x+C$