complex analysis - 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