Find the mathematical expectation of the given process?

Question

$$E[\cos(2\pi t+\theta)\sin(2\pi f(t+\tau)+\theta)]$$ where $$f=a+b\cos(\alpha)$$ and $\alpha$ is uniformly distributed on the interval (0,$\pi$). $a$ and $b$ are constants

My approach: $$=\cos(2\pi t+\theta)E[\sin(2\pi f(t+\tau)+\theta)]$$ $$=\cos(2\pi t+\theta)E[\sin(2\pi (a+b\cos(\alpha))(t+\tau)+\theta)]$$ $$=\cos(2\pi t+\theta)E[\sin(2\pi a(t+\tau)+2\pi b\cos(\alpha)(t+\tau)+\theta)]$$ $$=\cos(2\pi t+\theta)\int\limits_{0}^{\pi} \sin(2\pi a(t+\tau)+2\pi b\cos(\alpha)(t+\tau)+\theta)\frac{1}{\pi}d\alpha$$ $$\bigg/ 2\pi a(t+\tau)+2\pi b\cos(\alpha)(t+\tau)+\theta=x /d \bigg/$$ $$\bigg/d\alpha=-\frac{dx}{2\pi b\sin(\alpha)}\bigg/$$ $$=\cos(2\pi t+\theta)\frac{1}{2\pi^2 b\sin(\alpha)}\Big[\cos( 2\pi a(t+\tau)- 2\pi b(t+\tau)+\theta)-\cos( 2\pi a(t+\tau)+ 2\pi b(t+\tau)+\theta)\Big]$$ after applying: $$\cos x-\cos y = -2\sin\frac{x+y}{2}\sin\frac{x-y}{2}$$ I get: $$=\frac{\sin(2\pi a(t+\tau)+\theta)\sin(2\pi b(t+\tau))}{\pi^2b \sin(\alpha)}$$
Am I wrong or is this correct? For me the solution looks a little bit long and I was expecting a simpler solution. Maybe there are some properties I have overseen. Any help would be much appreciated.

Answer 1

Starting with the integral part of your 4th line.

$\int\limits_{0}^{\pi} \sin(2\pi a(t+\tau)+2\pi b\cos(\alpha)(t+\tau)+\theta)d\alpha$ $=\int\limits_{0}^{\pi} \sin(A + B cos(\alpha))d\alpha$ $=\pi \sin(A)J_0(B)$

With $J_0$ being a Bessel function of the first kind. I cheated and used Wolfram alpha to do the integral, but it is pretty much in the standard form for such an integral. So putting it all together.

$E[\cos(2\pi t+\theta)\sin(2\pi f(t+\tau)+\theta)]= \cos(2\pi t+\theta)\sin(2\pi a(t+\tau)+\theta)J_0(2\pi b(t+\tau))$