${\displaystyle \underset{0}{\overset{1}{\int }}x{\sin }^{–1}xdx=}$

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

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

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

                         $=\frac{\pi }{4}-\frac{1}{2}{\left(Si{n}^{-1}x\right)}_0^1+\frac{1}{2}\times \frac{\pi \times 1^2}{4}$

                          $=\frac{\pi }{8}\left(\because {\displaystyle \underset{0}{\overset{a}{\int }}\sqrt{a^2-x^2}}dx=\frac{\pi a^2}{4}\right)$