Integrate the function ( ^ -1 x)^2 - Vidyadip

Question

Integrate the function $(\sin^{-1}x)^2$

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Answer

Putting $x = \sin \theta $
$\Rightarrow dx = \cos \theta d\theta $
$\therefore \int {{{\left( {{{\sin }^{ - 1}}x} \right)}^2}} dx$
$= \int {{\theta ^2}\cos \theta } d\theta $
$[$Applying product rule$]$
$ = {\theta ^2}\sin \theta - \int {2\theta \sin \theta d\theta } $
$= {\theta ^2}\sin \theta - 2\int {\theta \sin \theta d\theta }$
$[$Again applying product rule$]$
$= {\theta ^2}\sin \theta - 2\left[ {\theta \left( { - \cos \theta } \right) - \int {1.\left( { - \cos \theta } \right)d\theta } } \right]$
$= {\theta ^2}\sin \theta + 2\theta \cos \theta - 2\int {\cos \theta d\theta } $
$= {\theta ^2}\sin \theta + 2\theta \cos \theta - 2\sin \theta + c$
$= x{\left( {{{\sin }^{ - 1}}x} \right)^2} + 2\sqrt {1 - {x^2}} {\sin ^{ - 1}}x - 2x + c$

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