Answer
$$\frac{\pi}{2}$$
Work Step by Step
We integrate as follows:
\begin{aligned} \int_{0}^{\pi} \sin ^{2} m x d x &=\frac{1}{m} \int_{0}^{m \pi} \sin ^{2} u d u \\ &=\frac{1}{m} \int_{0}^{m \pi}\left(\frac{1-\cos 2 u}{2}\right) d u \\ &=\left.\frac{1}{m}\left(\frac{u}{2}-\frac{\sin 2 u}{4}\right)\right|_{0} ^{m \pi} \\ &=\frac{1}{m}\left[\frac{m \pi}{2}-\frac{\sin 2 m \pi}{4}-0\right]\\
&=\frac{\pi}{2} \end{aligned}
