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.

Define $\displaystyle I(a,b)=\int_{-\infty}^{\infty} \frac{\sin^2ax\cos bx}{x^2} {\rm d}x \displaystyle$, and take Laplace transform of $I(a,b)$ with respect to $a$ and $b$, with Laplace domain variables $s$ and $t$, respectively:
\begin{align}
\mathcal{L}_{a\rightarrow s,b\rightarrow t}\{I(a,b)\}&=\int_{-\infty}^{\infty} \frac{2t}{(t^2+x^2)(s^3+4sx^2)} {\rm d}x\\
&=\frac{2\pi}{s^2(s+2t)}
\end{align}
Now taking inverse Laplace with respect to $t$ and then with respect to $s$ we obtain
\begin{align}
\mathcal{L}^{-1}_{s\rightarrow a}\Big\{\mathcal{L}^{-1}_{t\rightarrow b}\{\frac{2\pi}{s^2(s+2t)}\}\Big\}&=\mathcal{L}^{-1}_{s\rightarrow a}\{\pi\frac{e^{-\frac{bs}{2}}}{s^2}\}\\
&=\pi\Big(a-\frac{b}{2}\Big)H(a-\frac{b}{2}),
\end{align}
where $H(x)$ is the Heaviside function. Therefore for $a>\frac{b}{2}$we have $$I(a,b)=\int_{-\infty}^{\infty} \frac{\sin^2ax\cos bx}{x^2} {\rm d}x=\pi\Big(a-\frac{b}{2}\Big)$$ and for $a\leq\frac{b}{2}$ 
$$I(a,b)=\int_{-\infty}^{\infty} \frac{\sin^2ax\cos bx}{x^2} {\rm d}x=0.$$
This hence implies that 
$$\int_{-\infty}^{\infty} \frac{\sin^2x\cos x}{x^2} {\rm d}x=\pi\Big(1-\frac12\Big)=\frac{\pi}{2}.$$
