Integral $\int_{-\infty}^{\infty} \frac{\sin^2x}{x^2} e^{ix}dx$ How do I determine the value of this integral?
$$\int_{-\infty}^{\infty} \frac{\sin^2x}{x^2} e^{ix}dx$$
Plugging in Euler's identity gives
$$\int_{-\infty}^{\infty} \frac{i\sin^3x}{x^2}dx + \int_{-\infty}^{\infty} \frac{\sin^2x \cos x}{x^2}dx$$
and since $\dfrac{i\sin^3x}{x^2}$ is an odd function, all that is left is
$$\int_{-\infty}^{\infty} \frac{\sin^2x \cos x}{x^2}dx$$
at which point I am stuck.
I feel I am not even going the right direction, anybody willing to help?
Thanks in advance.

The integral can be evaluate by converting it to a double integral first.
Let
$$I = \int_{-\infty}^\infty \frac{\sin^2 x \cos x}{x^2} \, dx = 2 \int_0^\infty \frac{\sin^2 x \cos x}{x^2} \, dx.$$
Noting that
$$\sin^2 x \cos x = \frac{1}{4} (\cos x - \cos 3x),$$
yields
$$I = \frac{1}{2} \int_0^\infty \frac{\cos x - \cos 3x}{x^2} \, dx.$$
Now, by observing that
$$\int_1^3 \sin (tx) \, dt = \frac{\cos x - \cos 3x}{x},$$
the integral for $I$ can be rewritten as
$$I = \frac{1}{2} \int_0^\infty \int_1^3 \frac{\sin (xt)}{x} \, dt \, dx,$$
or
$$I = \frac{1}{2} \int_1^3 \int_0^\infty \frac{\sin (xt)}{x} \, dx \, dt,$$
after the order of integration has been changed. 
Enforcing a substitution of $x \mapsto x/t$ leads to
$$I = \frac{1}{2} \int_1^3 \int_0^\infty \frac{\sin x}{x} \, dx \, dt.$$
Making use of the well-known result of
$$\int_0^\infty \frac{\sin x}{x} \, dx = \frac{\pi}{2},$$
one has
$$I = \frac{\pi}{4} \int_1^3 dt,$$
or
$$\int_{-\infty}^\infty \frac{\sin^2 x \cos x}{x^2} \, dx = \frac{\pi}{2},$$
as expected.
