Answer
$$\eqalign{
& \int {{{\sin }^2}x} dx = \frac{1}{2}x - \frac{1}{4}\sin 2x + C \cr
& \int {{{\cos }^2}x} dx = \frac{1}{2}x + \frac{1}{4}\sin 2x + C \cr} $$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^2}x} dx \cr
& {\text{Use the identity }}{\sin ^2}x = \frac{{1 - \cos 2x}}{2} \cr
& \int {{{\sin }^2}x} dx = \int {\frac{{1 - \cos 2x}}{2}} dx \cr
& {\text{ }} = \int {\left( {\frac{1}{2} - \frac{1}{2}\cos 2x} \right)} dx \cr
& {\text{Integrating}} \cr
& {\text{ }} = \frac{1}{2}x - \frac{1}{4}\sin 2x + C \cr
& \cr
& \int {{{\cos }^2}x} dx \cr
& {\text{Use the identity }}{\cos ^2}x = \frac{{1 + \cos 2x}}{2} \cr
& \int {{{\cos }^2}x} dx = \int {\frac{{1 + \cos 2x}}{2}} dx \cr
& {\text{ }} = \int {\left( {\frac{1}{2} + \frac{1}{2}\cos 2x} \right)} dx \cr
& {\text{Integrating}} \cr
& {\text{ }} = \frac{1}{2}x + \frac{1}{4}\sin 2x + C \cr} $$
