I'm trying to show that
$$\int_{0}^{2\pi}\cos(n\theta-2\sin\theta)d\theta = 2\pi\sum_{r=0}^{\infty}\frac{(-1)^n}{r!(n+r)!}$$
The question hints that I should consider
$$e^{z-z^{-1}}$$
but I don't see how to get to the answer. Considering the nature of the integral, the unit circle seems like a sensible contour. The integrand has only one pole in the unit circle (at zero), which has residue $-1$. So:
$$\int e^{z-z^{-1}}dz=-2\pi i$$
Switching to $\theta$:
$$\int_0^{2\pi}ie^{i\theta}e^{e^{i\theta}-e^{-i\theta}}d\theta=-2\pi i$$
$$\int_0^{2\pi}e^{2\sin\theta+i\theta}d\theta=-2\pi$$
That integrand looks similar, but doesn't match up with the required integral, and I don't see how to get them to match. More troublesome, the required infinite sum on the right hand side is missing.
I've considered switching to a contour of radius $n$, but that just gives me $(2n+1)\pi$ on the right-hand side, and I still can't get it to match on the left. How do I solve this?
EDIT: Infinite sum corrected.