Suppose that $f$ is a smooth function which satisfies in the following condition
$$f(\pi+x)=f(x)\space \space and \space f(\pi-x)=f(x)$$ then, if the following integral exists
$$\int_0^\infty\frac{\sin^2x}{x^2}f(x)$$ We have the following nice equalities
$$\int_0^\infty\frac{\sin^2x}{x^2}f(x)=\int_0^{\infty}\frac{\sin x}{x}f(x)=\int_0^{\pi/2}f(x)dx$$
I proved the following formula as a part of my Bachelor project which extend the Lobachevsky's work that if we have $f(\pi+x)=f(x)\space \space and \space f(\pi-x)=f(x)$ then
$$\int_0^\infty\frac{\sin^4x}{x^4}f(x)=\int_0^{\pi/2}f(x)dx+\frac{2}{3}\int_0^{\pi/2}\sin^2 x\,f(x)dx$$
and I gave a method for finding an explicit formula for higher degree
$$\int_0^\infty\frac{\sin^{2n}x}{x^{2n}}f(x)$$