I am trying to solve:
$$ \int^{2 \pi}_{0} e^{cos \theta} cos(\theta -sin(\theta))d\theta $$
I tried to do the following substitution:
$ z = e^{i \theta } \rightarrow \theta =-i ln(z)$
$ dz = i e^{i \theta} d\theta = izd\theta $
$cos\theta = \frac{z+\frac{1}{z}}{{2}}$
$sin\theta = \frac{z-\frac{1}{z}}{{2i}}$
and I got:
$$ \int e^{z/2}e^{1/{2z}}cos(-i ln(z)-\frac{z-\frac{1}{z}}{{2i}})dz$$
I dont think this will take me anywhere
Is there a better aproach?
To help you get started, let $$A = \int_{0}^{2\pi}e^{\cos \theta}\cos(\theta-\sin \theta)\,d\theta,$$ and $$B = \int_{0}^{2\pi}e^{\cos \theta}\sin(\theta-\sin \theta)\,d\theta.$$
Obviously, both $A$ and $B$ are real numbers. Also, \begin{align}A-iB &= \int_{0}^{2\pi}e^{\cos \theta}[\cos(\theta-\sin \theta)-i\sin(\theta-\sin \theta)]\,d\theta \\ &= \int_{0}^{2\pi}e^{\cos \theta}e^{-i(\theta-\sin \theta)}\,d\theta \\ &= \int_{0}^{2\pi}e^{\cos \theta+i\sin\theta}e^{-i\theta}\,d\theta \\ &= \int_{0}^{2\pi}e^{e^{i\theta}}e^{-i\theta}\,d\theta \\ &= \oint_{C}\dfrac{e^{z}}{iz^2}\,dz \end{align} where $C$ is the curve parameterized by $z = e^{i\theta}$ over $\theta \in [0,2\pi]$. Can you take it from here?