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