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