Answer
$$y = \frac{1}{2}x - \frac{1}{4}\sin 2x$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = {\sin ^2}x,{\text{ }}\left( {0,0} \right) \cr
& {\text{Separate the variables}} \cr
& dy = {\sin ^2}xdx \cr
& {\text{Integrate both sides}} \cr
& \int {dy} = \int {{{\sin }^2}x} dx \cr
& y = \int {\frac{{1 - \cos 2x}}{2}} dx \cr
& y = \int {\left( {\frac{1}{2} - \frac{{\cos 2x\left( 2 \right)}}{4}} \right)} dx \cr
& y = \frac{1}{2}x - \frac{1}{4}\sin 2x + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}\left( {0,0} \right) \cr
& 0 = \frac{1}{2}\left( 0 \right) - \frac{1}{4}\sin 2\left( 0 \right) + C \cr
& C = 0 \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = \frac{1}{2}x - \frac{1}{4}\sin 2x \cr
& \cr
& {\text{Graph}} \cr} $$
