Answer
$$\frac{{{{\left( {{{\sin }^{ - 1}}x} \right)}^3}}}{3} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\left( {{{\sin }^{ - 1}}x} \right)}^2}dx}}{{\sqrt {1 - {x^2}} }}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = {\sin ^{ - 1}}x,{\text{ so that }}du = \frac{1}{{\sqrt {1 - {x^2}} }}dx \cr
& {\text{then}} \cr
& \int {\frac{{{{\left( {{{\sin }^{ - 1}}x} \right)}^2}dx}}{{\sqrt {1 - {x^2}} }}} = \int {{u^2}} du \cr
& {\text{integrate by the power rule}} \cr
& = \frac{{{u^3}}}{3} + C \cr
& {\text{write in terms of }}x;{\text{ replace }}{\sin ^{ - 1}}x{\text{ for }}u \cr
& = \frac{{{{\left( {{{\sin }^{ - 1}}x} \right)}^3}}}{3} + C \cr} $$
