calculate the following integral $\int_{0}^{2\pi}e^{\cos(t)}\cos(nt-\sin(t))dt$ Calculate the following integral $$\int_{0}^{2\pi}e^{\cos(t)}\cos(nt-\sin(t))dt$$
I thought maybe setting back to $z$ and trying to calculate with Residue Theorem but it didn't get me anywhere
 A: Well it can work by the Residue theorem, I assumed $n$ is a non negative integer
Firstly, using the exponential definition of $\cos \left(x\right)$ we will get that our integral is equal to: $$ Re\left(\int _0^{2\pi }\:e^{\cos \left(t\right)}e^{int-i\sin \left(t\right)}dt\right)$$
Using Euler's identity we get
$$Re\left(\int _0^{2\pi }\:e^{e^{-it}}e^{it\left(n-1\right)}e^{it}dt\right)$$
now letting $z=e^{it}$ we get contour integral of $$-iexp\left(\frac{1}{z}\right)z^{n-1}$$ where $\left|z\right|=1$
using Laurent expansion of $$\left(e^{\frac{1}{z}}z^{n-1}\right)=\sum _{k=0}^{\infty }\left(\frac{z^{n-1}}{k!z^k}\right)$$
By the residue theorem our contour integral $=2\pi i c_{-1}$ where $c_{-1}$ is the coefficient of $\frac{1}{z}$ term in our Laurent expansion
At $k=n$ we get $\frac{1}{z}$ term with cofficent of $\frac{1}{n!}$
Thus our integral is equal to $$\frac{2\pi }{n!}$$
if n was negative integer the function would be entire thus the integral would equal to $0$. You can easily prove that using the property of  $$\cos \left(-t\right)=\cos \left(t\right)$$
