Question

Solve

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

Answer 1

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

Put ${\cos }^{-1}x=t=1x=\cos t$

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

$-\int t\cos tdt$

$-t\int \cos tdt-\int \int \cos tdt$

$\Rightarrow \cos t-t\sin t$

$x-{\cos }^{-1}x\sqrt{1-x^2}$