Answer
$$x{\left( {{{\sin }^{ - 1}}x} \right)^2} + 2{\sin ^{ - 1}}x\sqrt {1 - {x^2}} - 2x + C$$
Work Step by Step
$$\eqalign{
& {\int {\left( {{{\sin }^{ - 1}}x} \right)} ^2}dx \cr
& {\text{let }}t = {\sin ^{ - 1}}x,\,\,\,\,\,\,\sin t = \sin \left( {{{\sin }^{ - 1}}x} \right)\,\,\,\, \to \,\,\,\,\sin t = x,\,\,\,\,\,\,\cos tdt = dx \cr
& {\text{Write the integrand in terms of }}t \cr
& {\int {\left( {{{\sin }^{ - 1}}x} \right)} ^2}dx = \int {{t^2}} \cos tdt \cr
& \cr
& {\text{Using integration by parts method }} \cr
& \,\,\,\,\,\,\,{\text{Let }}u = {t^2},\,\,\,\,\,\,\,du = 2tdt,\,\,\,\,\,\,dv = \cos tdt,\,\,\,\,v = \sin t \cr
& {\text{Integration by parts then gives}} \cr
& = {t^2}\sin t - \int {\sin t\left( {2tdt} \right)} \cr
& = {t^2}\sin t - \int {2t\sin tdt} \cr
& \cr
& {\text{Integrate by parts again}} \cr
& \,\,\,\,\,{\text{Let }}u = 2t,\,\,\,\,du = 2dt,\,\,\,\,dv = \sin tdt,\,\,\,\,v = - \cos t \cr
& = {t^2}\sin t - \left( { - 2t\cos t - \int {\left( { - \cos t} \right)\left( {2dt} \right)} } \right) \cr
& = {t^2}\sin t + 2t\cos t - \int {2\cos tdt} \cr
& {\text{Integrating}} \cr
& = {t^2}\sin t + 2t\cos t - 2\sin t + C \cr
& \cr
& {\text{Write in terms of }}x \cr
& = {\left( {{{\sin }^{ - 1}}x} \right)^2}\sin \left( {{{\sin }^{ - 1}}x} \right) + 2\left( {{{\sin }^{ - 1}}x} \right)\cos \left( {{{\sin }^{ - 1}}x} \right) - 2\sin \left( {{{\sin }^{ - 1}}x} \right) + C \cr
& {\text{Simplify}} \cr
& = x{\left( {{{\sin }^{ - 1}}x} \right)^2} + 2{\sin ^{ - 1}}x\sqrt {1 - {x^2}} - 2x + C \cr} $$
