Answer
$$\frac{{{{\left( {{{\sin }^{ - 1}}x} \right)}^2}}}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}} dx \cr
& {\text{set }}u = {\sin ^{ - 1}}x{\text{ then }}du = \frac{1}{{\sqrt {1 - {x^2}} }} \cr
& {\text{use the change of variable}} \cr
& \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}} dx = \int u du \cr
& {\text{integrate}} \cr
& = \frac{{{u^2}}}{2} + C \cr
& {\text{replace }}u = {\sin ^{ - 1}}x \cr
& = \frac{{{{\left( {{{\sin }^{ - 1}}x} \right)}^2}}}{2} + C \cr} $$
