Integrals — Maths STD 12 Science — Question

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

$= \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 } $

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

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