$$\int_{-1/2}^{1/2}\sin^2(2^{j-1}\pi f)\prod_{i=0}^{j-2} \cos^2(2^i\pi f)df$$
I've tried simplifying the integrand, but I can't get to a point where I can evaluate the integral.
I know $\prod_{i=0}^{j-2} \cos^2(2^i\pi f) = \frac{\sin^2(2^{j-1}\pi f)}{2^{2j-2} \sin^2(\pi f)}$ if the denominator is suitably nonzero.
Which yields $$\int_{-1/2}^{1/2}\frac{\sin^4(2^{j-1}\pi f)}{2^{2j-2} \sin^2(\pi f)}df$$
but from here I keep getting stuck. Any help?
Hint: Use the identity $\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ and the fact that: $$ \forall n\in\mathbb{Z},\qquad \int_{-1/2}^{1/2}e^{2\pi niz}\,dz = \delta(n) $$ to prove that: $$ \forall n\in\mathbb{N},\qquad \int_{-1/2}^{1/2}\frac{\sin^4(\pi n x)}{\sin^2(\pi x)}\,dx = \frac{n}{2}.$$