Contour Integration of $\sin^2(x)/(1+x^2)$ How should I calculate this integral 
$$\int\limits_{-\infty}^\infty\frac{\sin^2x}{(1+x^2)}\,dx\quad?$$ 
I have tried forming an indented semicircle in the upper half complex plane using the residue theorem and I tried to integrate along a curve that went around the complex plane and circled the positive real axis (since the integrand is even). Nothing has worked out for me. 

First we simplify the integral the following way using the simple trigonometric identity $\sin^2 x = \frac 12 - \frac 12 \cos{2x}$: $$\begin{align*}\int_{-\infty}^{\infty} \frac{\sin^2 x}{1+x^2}\,dx &= \frac 12 \int_{-\infty}^{\infty} \frac{dx}{1+x^2} - \frac 12 \int_{-\infty}^{\infty} \frac{\cos {2x}}{1+x^2}\,dx\\&= \frac{\pi}{2} - \frac 12 \int_{-\infty}^{\infty} \frac{\cos {2x}}{1+x^2}\,dx\end{align*}$$
Now to solve this second integral we note that it is equivalent to $\int_{-\infty}^{\infty} \frac{e^{2ix}}{1+x^2}\,dx$ as the imaginary part is odd so it's integral evaluates to 0. We now consider the semi circle contour integral of radius $R$ as $R \to \infty$ in the top half plane for this function traveling in an anti-clockwise direction, encompassing the pole at $z = i$.
$$\oint \frac{e^{2iz}}{1+z^2}\,dz = \int_{-R}^{R}\frac{e^{2iz}}{1+z^2}\,dz + \int_{\gamma} \frac{e^{2iz}}{1+z^2}\,dz$$ where $\gamma$ is the semi-circle arc contour. By Cauchy's integral formula we have $$\oint \frac{e^{2iz}}{1+z^2}\,dz = \oint \frac{\frac{e^{2iz}}{z+i}}{z-i}\,dz = 2\pi i \frac{e^{-2}}{2i} = {\pi \over e^2}$$
Now we look at the semi-circle arc contour. $$\lvert z^2 + 1 - 1 \rvert \leq \lvert z^2 + 1 \rvert + 1 \implies \lvert z^2 + 1 \rvert \geq R^2 - 1\\\left\lvert e^{2iz}\right\rvert \leq 1$$
Hence by the estimation lemma:
$$\left\lvert \int_{\gamma} \frac{e^{2iz}}{1+z^2}\,dz \right\rvert \leq \frac{\pi R}{R^2 - 1}$$
Finally, taking the limit as $R \to \infty$ gives us
$$\int_{-\infty}^{\infty} \frac{e^{2ix}}{1+x^2}\,dx = {\pi \over e^2} \\ \implies \int_{-\infty}^{\infty} \frac{\sin^2 x}{1+x^2}\,dx = \frac{\pi}{2} - \frac{\pi}{2e^2}$$ 
