$\begin{align}  & \int{\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}dx} \\ & \text{Let }u={{\sin }^{-1}}x,\text{ }\frac{du}{dx}=\frac{1}{\sqrt{1-{{x}^{2}}}};\text{ }dx=\sqrt{1-{{x}^{2}}}du \\ & \int{\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}dx}=\int{\frac{u}{\sqrt{1-{{x}^{2}}}}(\sqrt{1-{{x}^{2}}}du)}=\int{udu} \\ & \int{\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}=\frac{{{u}^{2}}}{2}+C=\frac{{{({{\sin }^{-1}}x)}^{2}}}{2}+C \\\end{align}$