${\left( {{{\sin }^{ - 1}}x} \right)^2}$

Let $I = \int {{{\left( {{{\sin }^{ - 1}}x} \right)}^2}} dx$

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

$\therefore$ $I = \int {{\theta ^2}\cos \theta d\theta }$

$= {\theta ^2}\int {\cos \theta \,d\theta } - \int {\left( {\cfrac{d}{{d\theta }}\left( {{\theta ^2}} \right) \cdot \int {\cos \theta d\theta } } \right)} d\theta$

$= {\theta ^2}\sin \theta - \int {2\theta \sin \theta d\theta = {\theta ^2}\sin \theta - 2\int {\theta \sin \theta d\theta } + C}$

$= {\theta ^2}\sin \theta - 2\left[ {\theta \left( { - \cos \theta } \right) - \int {\left( 1 \right)\left( { - \cos \theta } \right)d\theta } } \right] + C$

$= {\theta ^2}\sin \theta + 2\theta \cos \theta - 2\int {\cos \theta \,d\theta } + C$

$= {\theta ^2}\sin \theta + 2\theta \sqrt {1 - {{\sin }^2}\theta } - 2\sin \theta + C$

$= x{\left( {{{\sin }^{ - 1}}x} \right)^2} + 2{\sin ^{ - 1}}x\sqrt {1 - {x^2}} - 2x + C$