I was surprised by the value of an integral I saw in Wolfram Alpha so I wanted to prove it on my own. This is the following integral : $$ \displaystyle I=\int_{0}^{+\infty}\frac{\sin^2\left(t\right)}{1+t^2}\text{d}t=\frac{\pi}{2e}\text{sinh}\left(1\right)$$

Is there an easy way to show this equality ?

In order to prove it, my idea was to introduce its twin integral $ \displaystyle J=\int_{0}^{+\infty}\frac{\cos^2\left(t\right)}{1+t^2}\text{d}t$. Then we have two results : $$I+J=\frac{\pi}{2}$$ $$I-J=\int_{0}^{+\infty}\frac{\cos\left(2t\right)}{1+t^2}\text{d}t$$ I would like to show that for $w>0$ $$\int_{0}^{+\infty}\frac{\cos\left(wt\right)}{1+t^2}\text{d}t=\frac{\pi}{2}e^{-t}$$ I've searched for a differential equation with its derivative $\displaystyle -\int_{0}^{+\infty}\frac{t\sin\left(wt\right)}{1+t^2}\text{d}t$ but it doesnt seem that nice even when I integrate by parts... Any ideas ?