Evaluating $\int_0^{2 \pi} e^{\cos x} \cos (nx - \sin x) \,dx$ using complex analysis I'm taking a complex analysis course and doing some practice computing residues & evaluating integrals. I pulled out an old book called  "The Cauchy Method of Residues: Theory and Applications, Volume I" 
On page 196-197, there are some interesting integrals to evaluate. I'm at 5.4.3.10.: I was able to do question 1, but was stumped at how to even begin with question 2:

Evaluate the integral
  $$\int_0^{2 \pi} e^{\cos x} \cos (nx - \sin x) \,dx ,$$
  where $n$ is an natural number.

The answer is simply $\frac{2 \pi}{n}$. Any hints?

Note
$$\int_0^{2\pi} e^{\cos x}\cos(nx - \sin x)\, dx = \operatorname{Re} \int_0^{2\pi} e^{\cos x + i(nx - \sin x)}\, dx = \operatorname{Re} \int_0^{2\pi} e^{e^{-ix}} e^{inx}\, dx.$$
Using the parametrization $z = e^{-ix}$, $0 \le x \le 2\pi$ for the unit circle $|z| = 1$, we have 
$$\int_0^{2\pi} e^{e^{-ix}}e^{inx}\, dx = \frac{1}{i}\int_{|z| = 1} e^{z} \frac{dz}{z^{n+1}}.$$
Now show that this contour integral is $\frac{2\pi}{n!}$.
