How to Apply Cauchy Integral Theorem I have a question about complex analysis. 
Let C be the unit circle z = ejθ described from θ = -π to θ = π where α is a real constant. I need to prove this equation:

I thought that somehow I should rearrange the integrant to apply cauchy integral theorem. Thus I need singularity offspring of it. 
Then I made variable displacement like this:

Now, I have the singularity but I'am stuck. How to go on with that?

We have 
\begin{align}
\int_0^\pi e^{a\cos \theta} \cos(a\sin \theta)\, d\theta &= \frac{1}{2}\int_{-\pi}^{\pi} e^{a\cos \theta} \cos(a\sin \theta)\, d\theta \\
&= \frac{1}{2}\operatorname{Re} \int_{-\pi}^{\pi} e^{ae^{i\theta}}\, d\theta\\
&= \frac{1}{2}\operatorname{Re} \oint_C e^{az} \frac{dz}{iz} \tag{*}\\
\end{align}
Since $e^{az}/i$ is analytic inside and on $C$, and $0$ lies inside $C$, Cauchy's integral formula gives 
$$\oint_C e^{az} \frac{dz}{iz} = 2\pi i\cdot \frac{e^{az}}{i}\bigg|_{z = 0} = 2\pi.$$
Hence, by $(*)$, 
$$\int_0^\pi e^{a\cos \theta}\cos(a\sin \theta)\, d\theta = \pi.$$
