Answer
$$y = \sin \left( {{x^2} + C} \right)$$
Work Step by Step
$$\eqalign{
& y' = 2x\sqrt {1 - {y^2}} \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = 2x\sqrt {1 - {y^2}} \cr
& {\text{Separate the variables}} \cr
& \frac{{dy}}{{\sqrt {1 - {y^2}} }} = 2xdx \cr
& {\text{Integrate both sides}} \cr
& \int {\frac{{dy}}{{\sqrt {1 - {y^2}} }}} = \int {2x} dx \cr
& \arcsin y = {x^2} + C \cr
& {\text{Solve for }}y \cr
& \sin \left( {\arcsin y} \right) = \sin \left( {{x^2} + C} \right) \cr
& y = \sin \left( {{x^2} + C} \right) \cr} $$
