Improper integral of $\sin^2(x)/x^2$ evaluated via residues

I have come across another improper integral I wish to evaluate via residues.

The integral is:

$$\int_{-\infty}^\infty{\frac{\sin(x)^2}{x^2}}dx$$

$\sin(z)$ behaves in an uneasy way so I tried using the function $\frac{{e^{iz}}^2}{z^2}$ with a half circle on the upper complex plane with radius R and a half-circle of radius 1/R which arcs below $0$.

The problem is the small semi-circles integral does not go to $0$ and in fact doesn't exist.

What other types of contours or function substitutions should be used here?

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Is it $\,\sin^2x\,$ (headline) or $\,\sin x\,$ (message)? –  DonAntonio Oct 17 '12 at 5:00
$sin(x)^2$, I wrote the function down incorrectly but then edited it. –  Mike Oct 17 '12 at 5:02
Is that $\exp(iz^2)$ or $\exp(iz)^2=\exp(2iz)$? I don't see how they're linked with $\sin(x)^2$ anyway. –  Philippe Malot Oct 26 '13 at 18:28

Note that $\cos(2x)=1-2\sin(x)^2$, this suggest to consider the integral
$$\int_{C} \frac{ {\rm e}^{2 i z} - 1 }{ z^2} dz \,.$$
Yes but how does one get rid of the $\frac{1}{z^2}$ term? –  Mike Oct 17 '12 at 16:10