Improper integral: $\int_0^\infty \frac{\sin^4x}{x^2}dx$

The inequality $\left|\sin x\right|\leq\min(1,|x|)$ is enough to prove that the integral is converging.

Integration by parts and the well-known $\int_{0}^{+\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2}$ then give: $$\begin{eqnarray*} I = \int_{0}^{+\infty}\frac{\sin^4 x}{x^2}\,dx &=& \int_{0}^{+\infty}\frac{4\cos x \sin^3 x}{x}\,dx\\&=&\int_{0}^{+\infty}\frac{\sin(2x)-\frac{1}{2}\sin(4x)}{x}\,dx\\&=&\left(1-\frac{1}{2}\right)\int_{0}^{+\infty}\frac{\sin x}{x}\,dx=\color{red}{\frac{\pi}{4}}.\end{eqnarray*}$$