$\underset{0}{\overset{1}{\int }}x{\sin }^{-1}x dx=$

$\underset{0}{\overset{1}{\int }}x.{\sin }^{-1}xdx$

$=\underset{0}{\overset{1}{\int }}\left({\sin }^{-1}x\right).xdx$

$={\left(\left({\sin }^{-1}x\right).\frac{{\displaystyle x^2}}{{\displaystyle 2}}\right)}_0^1-\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-x^2}}\frac{x^2}{2}dx$

$=\left({\sin }^{-1}1.\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)-0+\frac{1}{2}\underset{0}{\overset{1}{\int }}\frac{(1-x^2)-1}{\sqrt{1-x^2}}dx$

$=\frac{\pi }{4}+\frac{1}{2}\underset{0}{\overset{1}{\int }}\left(\sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}}\right)dx$

$=\frac{\pi }{4}+\frac{1}{2}{\left(\frac{{\displaystyle x}}{{\displaystyle 2}}\sqrt{1-x^2}+\frac{{\displaystyle 1}}{{\displaystyle 2}}{\sin }^{-1}x\right)}_0^1-\frac{1}{2}{\left({\sin }^{-1}x\right)}_0^1$

$=\frac{\pi }{4}+\frac{1}{2}\left(\left(0+\frac{{\displaystyle 1}}{{\displaystyle 2}}\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)-0\right)-\frac{1}{2}\frac{\pi }{2}$

$=\frac{\pi }{4}+\frac{\pi }{8}-\frac{\pi }{4}=\frac{\pi }{8}$