Not that tough

Note that $\begin{aligned} \int_0^\infty \frac{\sin^{2n}x}{m^2\pi^2 - x^2}\,dx & = \; \frac12\int_{-\infty}^\infty \frac{\sin^{2n}x}{m^2\pi^2 - x^2}\,dx \; = \; \frac{1}{4m\pi}\int_{-\infty}^\infty \left(\frac{1}{m\pi-x} + \frac{1}{m\pi+x}\right)\sin^{2n}x\,dx \\ & = \; \frac{1}{4m\pi}\left(-\int_{-\infty}^\infty \frac{\sin^{2n}x}{x}\,dx + \int_{-\infty}^\infty \frac{\sin^{2n}x}{x}\,dx\right) \; = \; 0 \end{aligned}$ for any integers $m,n \in \mathbb{N}$ . The desired integral splits, using partial fractions, in to a sum of integrals of the above type, and hence is equal to $\boxed{0}$ as well.