Use contour integration to calculate the integral $\int_{-\infty}^{+\infty} \frac{\sin^2(x)}{x^2+a^2}dx$ with $a>0$
For this I have a solution (where they expand $\sin^2(x) = \frac{1}{2}(1-\cos(2x))$ and then put $\sin^2(x) = \frac{1}{2}\Re(1-e^{2it})$. But when I first attempted it I tried it with the function $$f(z) = \frac{\sin^2(z)}{z^2+a^2}.$$
I know that the poles are $+ia,-ia$. Consider a contour integral which is a halfcircle on the upperhalf plane, $\gamma_R = [-R,R] \cup C_R$.
With the residue formula $\int_{\gamma_R}f(z) dz$ gives me $\frac{\pi}{a}(\sinh(a))^2$.
But I'm having troubles with getting $\int_{C_R} f(z)dz \leq 0$. I first tried this \begin{align} \big|\int_{C_R} f(z)dz\big| &\leq \int_{C_R}\big|\frac{\sin^2(x)}{x^2+a^2}\big|dz\\ &\leq \int_{C_R}\big|\frac{1}{x^2+a^2}\big|dz \end{align}
But apparently $sin(z)\nleq 1$ for $z \in \mathbb{C}$.
My question is, is it possible to find an estimate with this complex function so that I can get $\int_{C_R} f(z)dz \leq 0$? Or do I have to use the formula $\sin^2(x) = \frac{1}{2}(1-\cos(2x))$?