The value of the integral ${\int }_0^{1/2} \frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx$, is:

Let $\mathrm{I}={\int }_0^{1/2} \frac{{\mathrm{xsin}}^{-1}\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}$

Put ${\sin }^{-1}x=\theta \Rightarrow x=\sin \theta$

Lower limit $x=0,\theta ={\sin }^{-1}(0)=0$

Upper limit $x=1/2,\theta ={\sin }^{-1}(1/2)=\frac{\pi }{6}$

$\begin{array}{l}I={\int }_0^{\pi /6} \theta \sin \theta d\theta \\ =-{\left.\theta \cos \theta \right|}_0^{\pi /6}-{\int }_0^{\pi /6}( -\cos \theta )d\theta \end{array}$

$\begin{array}{l}=-\frac{\pi }{6}\cdot \frac{\sqrt{3}}{2}+{\left.\sin \theta \right|}_0^{\pi /6}\\ =\frac{-\sqrt{3}\pi }{12}+\frac{1}{2}=\frac{1}{2}-\frac{\pi \sqrt{3}}{12}.\end{array}$