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

HINT: Use the relation 
$$\int_0^\infty  \left(\frac{\sin x}{x}\right)^n \mathrm{d}x
= \frac{\pi}{2^n (n-1)!}  
    \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} (n-2k)^{n-1}$$
You may find a proof here A sine integral $\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$.
