Need help with an integral. I am asked to show that
$\displaystyle \frac{1}{2\pi} \int_{0}^{2\pi} e^{2 \cos \theta} \cos \theta \, d\theta = 1 + \frac{1}{2!} + \frac{1}{2!3!} + \frac{1}{3!4!} + \cdots$
by considering $\displaystyle \int e^{z + \frac{1}{z}} \, dz$. I don't really know how to incorporate the hint, and any advice would be appreciated. Thanks!

Spelling out David's hint, if $z=e^{i\theta}$ then $dz=i\,e^{i\theta}\,d\theta=i(\cos\theta+i\,\sin\theta)\,d\theta$, and we have  $z+\displaystyle\frac1z=2\cos\theta$.
Now
$$\frac1{2\pi}\int_0^{2\pi}e^{2\cos\theta}\,\cos\theta\,d\theta = 
\frac1{2\pi i}\Im\oint e^{z+\frac1z}\,dz
$$
which can be evaluated by the residue theorem: look for the coefficient of $z^{-1}$ in the Laurent-series of $e^z\,e^{\frac1z}$.
