How to calculate the integral $$\int_{0}^{+ \infty } \frac{\sin^{2} x}{x^{2}}dx$$
I suppose we should start from the known one:$$\int_{0}^{+ \infty } \frac{\sin x}{x}dx = \pi /2$$

You van solve this using integration by parts: integrating $x^{-2}$ gives
$$
\int_0^\infty \frac{\sin^2(x)}{x^2}dx=-\lim\limits_{x\to \infty}\frac{\sin^2(x)}{x}+\lim\limits_{x\to 0}\frac{\sin^2(x)}{x}+\int_0^\infty \frac{2\sin(x)\cos(x)}{x}dx
$$
The first two limits are 0 and using the formula $\sin(2x)=2\sin(x)\cos(x)$, the integral simplifies to
$$
\int_{0}^\infty \frac{\sin^2(x)}{x^2}dx=\int_0^\infty \frac{\sin(2x)}{x}dx
$$
Which with the change of variable $u=2x$ and the kwon integral yields
$$
\int_{0}^\infty \frac{\sin^2(x)}{x^2}dx= \frac{\pi}{2}
$$
