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