Evaluate: ∫_0^π (sin 2 p x)/(sin x) d x, p ∈ N.

\[\begin{aligned} & \mathrm{I}=\int_{0}^{\pi} \frac{\sin 2 \mathrm{p} x}{\sin x} d x \\ & =\int_{0}^{\pi} \frac{\sin 2 \mathrm{p}(\pi-x)}{\sin (\pi-x)} d x \\ & \mathrm{I}=\int_{0}^{\pi} \frac{-\sin 2 \mathrm{p} x}{\sin x} d x \end{aligned} \] Adding, we get $\displaystyle 2 \mathrm{I}=0$ $\displaystyle \therefore \mathrm{I}=0$