Solving $\int _0^{2\pi }e^{cos\left(\theta \right)}\:cos\left(n\theta \right)d\theta $ seem to be lacking inspiration towards solving this $\int _0^{2\pi }e^{cos\left(\theta \right)}\:cos\left(n\theta \right)d\theta $. Perhaps using the obvious method, but then, how does one calculate the residue of zero?
Many thanks in advance for hints.
 A: To provide us more mathematical power, first rewrite into a contour around the unit circle, 
$$A_n:=\frac{1}{2i}\int_C e^{(1/2)(z+1/z)}\cdot \left(z^{n-1}+\frac{1}{z^{n+1}}\right)dz. $$ 
Expand into summations, 
$$A_n=\frac{1}{2i}\int_C \left(\sum_{k=0}^\infty \frac{1}{k!2^k} \left(z+\frac{1}{z}\right)^k z^{n-1}+\sum_{k=0}^\infty \frac{1}{k!2^k} \left(z+\frac{1}{z}\right)^k z^{-n-1}\right)dz. $$ 
Compute the constituent pieces; by symmetry,
$$\int_C \left(z+\frac{1}{z}\right)^k z^{\pm n-1} = 2\pi i \binom{k}{(n+k)/2}.$$
Hence, 
$$\frac{A_n}{2\pi}=\sum_{k=0}^\infty \frac{1}{k!2^k}\binom{k}{(n+k)/2}=\left\{\begin{matrix}\sum_{k=0}^\infty \frac{1}{(k+m)!(k-m)!4^k} &\text{where}&n=2m \\ \sum_{k=0}^\infty \frac{1}{2(k+m+1)!(k-m)!4^k} &\text{where}&n=2m+1 \end{matrix}\right.$$ 
These series don't converge to explict answers, hence the next best thing is to match these series with entire functions that are well known. Thus, mathematica produces "Bessel" functions evaluated at one. Specifically, 
$$A_n=2\pi I_n(1),$$ where $$I_n(z)=\sum_{k=0}^\infty \frac{1}{k!(n+k)!}\left(\frac{z}{2}\right)^{2k+n}. $$
