Prove $\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}$ I need to show that 
$$
\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}
$$
I have already derived the result $\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}$ using complex analysis, a result which I am supposed to start from.  Using a change of variable $ x \mapsto 2x $ :
$$
\int_0^\infty \frac{\sin^2(2x)}{x^2}dx = \pi
$$
Now using the identity $\sin^2(2x) = 4\sin^2x - 4\sin^4x $, we obtain
$$
\int_0^\infty \frac{\sin^2x - \sin^4x}{x^2}dx = \frac{\pi}{4}
$$
$$
\frac{\pi}{2} - \int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4}
$$
$$
\int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4}
$$
But I am now at a loss as to how to make $x^4$ appear at the denominator.  Any ideas appreciated.
Important: I must start from $ \int_0^\infty \frac{\sin^2x}{x^2}dx $, and use the change of variable and identity mentioned above

You mentioned that you used complex analysis to evaluate $\int_{0}^{\infty} \frac{\sin^{2}(x)}{x^{2}} \, \mathrm dx$.
We can also use complex analysis to evaluate $\int_{0}^{\infty} \frac{\sin^{4}(x) }{x^{4}} \, \mathrm dx$.
Using the trigonometric identity $ \displaystyle \sin^{4} x = \frac{1}{8} \Big(\cos 4x - 4 \cos 2x + 3 \Big)$, we get
$$ \begin{align} \int_{0}^{\infty} \frac{\sin^{4} x}{x^{4}} \, \mathrm dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^{4} x}{x^4} \, \mathrm dx \\ &= \frac{1}{16} \int_{-\infty}^{\infty} \Re \ \frac{e^{4ix}-4e^{2ix}+3}{x^{4}} \, \mathrm  dx \\ &= \frac{1}{16}\int_{-\infty}^{\infty} \Re \ \frac{e^{4ix}-4e^{2ix}+3+4ix}{x^{4}} \, \mathrm dx \\ &= \frac{1}{16} \, \Re \, \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3+4ix}{x^{4}} \, \mathrm dx. \end{align}$$
So let's integrate the function $$f(z) = \frac{e^{4iz}-4e^{2iz}+3+4iz}{z^{4}}$$ around a contour the consists of the real axis from $-R$ to $R$, $R>0$, and the upper half of the circle $|z|=R$. To avoid the simple pole at the origin, the contour needs to be indented at the origin.
Letting the radius of the indentation go to zero and $R \to \infty$, we get
$$ \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3+4ix}{x^{4}} \, \mathrm dx- i \pi \ \text{Res}[f(z),0] = 0,$$
where
$$ \begin{align} \operatorname{Res}[f(z),0] &= \lim_{z \to 0} \frac{e^{4iz}-4e^{2iz}+3+4iz}{z^{3}} \\ &= \lim_{z \to 0} \frac{-64ie^{4iz}+32ie^{2iz} }{6} \\ &= - \frac{16i}{3}. \end{align}$$
Therefore,
$$ \int_{0}^{\infty} \frac{\sin^{4} x}{x^{4}} \, dx = \frac{1}{16} \left(\frac{16 \pi}{3} \right)=\frac{\pi}{3} .$$

Technically, it wasn't necessary to add $4ix$ to the numerator.
For reasons explained here, the Cauchy principal value of $ \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3}{x^{4}} \, \mathrm dx $ exists even though $\frac{e^{4iz}-4e^{2iz}+3}{z^{4}}$ has a pole of order $3$ at the origin.
