Using Laurent series to evaluate integral Use the Laurent series of $e^{1/z}$ on the punctured plane $\mathbb{C} - \{0\}$ to compute $1/\pi \int_0^\pi e^{\cos\theta}\cos(\sin\theta-n\theta) d\theta$ for $n=0,1,2,...$ 
I have been trying to attempt this question from my complex analysis book and I am not sure where to even begin. All I know is that the Laurent series of $e^{1/z}$ is $f(z)=\sum_{n=0}^{\infty} \frac{1}{n!}\frac{1}{z^n}$, for all $|z|>0$. 
I have no idea how this relates the the integral though because there is no $e^{1/z}$ only a $e^{\cos\theta}$ in the integral. I thought maybe I have to put the $\cos\theta$ values into Taylor form but I'm not sure if that's correct.
Any help is greatly appreciated! Thanks :)

Let $\gamma$ be the unit circle, and consider the integral
$$I_m=\int_\gamma z^m\exp(1/z)\,dz$$
where $m$ is an integer. Writing $z=e^{it}$ gives
$$I_m=i\int_0^{2\pi}e^{im t}e^{\cos t-i\sin t}e^{it}\,dt
=i\int_0^{2\pi}e^{\cos t}e^{i((m+1)t-\sin t)}\,dt.$$
The imaginary part of this integral is
$$\int_0^{2\pi}e^{\sin t}\cos((m+1)t-\sin t)\,dt$$
which is more-or-less then integral you are after.
But looking at the Laurent series of $e^{1/z}$ and using calculus of
residues gives $I_m=0$ for $m\le -2$ and 
$$I+m=\frac{2\pi i}{(m+1)!}$$
for $m\ge-1$.
