$\begin{aligned} \int_{0}^{\frac{\pi}{2}} \sin ^{2} x d x &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}(1-\cos 2 x) d x=\frac{1}{2}\left[x-\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}} \\ &=\frac{1}{2}\left[\frac{\pi}{2}+0\right]=\frac{\pi}{4} \end{aligned}$