Integral $\int\frac{(\sin(x))^2}{x^2+1} dx$ I have no idea of variable changement to use or other to calculate this integral :
$$
\int_0^{\infty}\frac{(\sin(x))^2}{x^2+1}\,dx 
$$
Wolfram alpha gives me the result but really no idea ... 
I seek a calculus without residues theorem or other big theorems like it because I didn't learn them.. so just by part or by variable changement if possible sure.. 
Thanks in advance,
Shadock 
 A: Here it is a proof with the residue theorem. We have:
$$ I = \int_{0}^{+\infty}\frac{1-\cos(2x)}{2(1+x^2)}\,dx = \frac{\pi}{4}-\frac{1}{2}\int_{0}^{+\infty}\frac{\cos(2x)}{1+x^2}\,dx=\frac{\pi}{4}-\frac{1}{4}\int_{\mathbb{R}}\frac{\cos(2x)}{1+x^2},$$
but:
$$ \int_{\mathbb{R}}\frac{\cos(2x)}{1+x^2}\,dx = \Re\int_{-\infty}^{+\infty}\frac{e^{2iz}}{1+z^2}\,dz=\Re\left(2\pi i\cdot\operatorname{Res}\left(\frac{e^{2iz}}{1+z^2},z=i\right)\right),$$
so:
$$\int_{\mathbb{R}}\frac{\cos(2x)}{1+x^2}\,dx = \frac{\pi}{e^2}\tag{1} $$
and:
$$ I = \int_{0}^{+\infty}\frac{\sin^2 x}{1+x^2}\,dx = \color{red}{\frac{\pi}{4}\left(1-\frac{1}{e^2}\right)}=\frac{\pi}{2e}\sinh 1.$$
With a probabilistic argument, $(1)$ follows from the fact that the CF of the standard Cauchy distribution is $e^{-|t|}$. To prove it, it is enough to show that:
$$\int_{0}^{+\infty}e^{-t}\cos(nt)\,dx = \frac{1}{1+n^2}\tag{2}$$
holds by integration by parts.
A: I particularly like Jack's hint concerning the Laplace transform. Write
$$I(a)=\int_0^\infty \frac{\sin^2(ax)}{x^2+1}\,dx$$
Then,
$$\mathcal{L}\left\{I(a)\right\}=\int_0^\infty\int_0^\infty\frac{\sin^2(ax)}{x^2+1}e^{-as}\,da\,dx=\int_0^\infty\frac{2x^2}{(x^2+1)(s^3+4sx^2)}\,dx$$
