integrate (sin inverse x)whole square

Putting {tex}x = \sin \theta \Rightarrow dx = \cos \theta d\theta {/tex} 

{tex}\therefore \int {{{\left( {{{\sin }^{ - 1}}x} \right)}^2}} dx{/tex}

{tex}= \int {{\theta ^2}\cos \theta } d\theta {/tex}

[Applying product rule]

{tex} = {\theta ^2}\sin \theta - \int {2\theta \sin \theta d\theta } {/tex}

{tex}= {\theta ^2}\sin \theta - 2\int {\theta \sin \theta d\theta }{/tex}

[Again applying product rule]

{tex}= {\theta ^2}\sin \theta - 2\left[ {\theta \left( { - \cos \theta } \right) - \int {1.\left( { - \cos \theta } \right)d\theta } } \right]{/tex}

{tex}= {\theta ^2}\sin \theta + 2\theta \cos \theta - 2\int {\cos \theta d\theta } {/tex}

{tex}= {\theta ^2}\sin \theta + 2\theta \cos \theta - 2\sin \theta + c{/tex}

{tex}= x{\left( {{{\sin }^{ - 1}}x} \right)^2} + 2\sqrt {1 - {x^2}} {\sin ^{ - 1}}x - 2x + c{/tex}