Integration of $\int_0^\infty\frac{\sin^2x}{x^2}dx$ [duplicate]

I have been trying very hard to find the answer to the following integral$$\int_0^\infty\frac{\sin^2x}{x^2}dx$$ given that $$\int_0^\infty\frac{\sin x\cos x}{x}dx = \frac{\pi}{4}$$

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See here for general version. –  Cortizol Sep 5 '13 at 14:25
Use integration by parts (and the fact that $\sin^2x/x$ vanishes for $x \rightarrow 0$ and $x\rightarrow\infty$) $$\int_0^\infty\frac{\sin^2x}{x^2}dx = \int_0^\infty\frac{2\sin x\cos x}{x}dx= \frac{\pi}{2}$$