Solving the Integral without Cauchy Integral formula. I saw this integral in a competitive exam which I donot recall, but I found this problem in my notebook which has to be shown without using Cauchy integral theorem, Poisson integral formula over circle - 
$$\int_{0}^{2\pi} \frac{e^{\cos(\phi)} \cos(\sin \phi)}{5 - 4\cos(\theta - \phi)} d\phi = \frac{2\pi}{3} e^{\cos(\theta)} \cos(\sin(\theta))$$
and
$$\int_{0}^{2\pi} \frac{e^{\cos(\phi)} \sin(\sin \phi)}{5 - 4\cos(\theta - \phi)} d\phi = \frac{2\pi}{3} e^{\cos(\theta)} \sin(\sin(\theta))$$
I think these are two beautiful and smart integrals(by comparing the two integrals which just differ by a trigonometric function), but I have no idea how to start with?. There may be some great idea involved in here. 
I thought of multiplying $i$ to the second integral and adding to the first integral which reduces to a term $e^{e^{i\phi}}$ but I think that will not help much.
Any clever approaches to this problem?
 A: Hint using residue theorem (and a bit of sweat) we can extend your idea of multiplying $i$ to the second integral and adding...:
$$\int_{0}^{2\pi} \frac{e^{\cos{\phi}} \cos(\sin{\phi})}{5 - 4\cos(\theta - \phi)} d\phi + i\int_{0}^{2\pi} \frac{e^{\cos{\phi}} \sin(\sin{\phi})}{5 - 4\cos(\theta - \phi)} d\phi = \\
\int_{0}^{2\pi} \frac{e^{\cos{\phi}} \left(\cos({\sin \phi})+i\sin(\sin{\phi})\right)}{5 - 4\cos(\theta - \phi)} d\phi =\\
\int_{0}^{2\pi} \frac{e^{\cos{\phi}} e^{i\sin{\phi}}}{5 - 4\cos(\theta - \phi)} d\phi =
\int_{0}^{2\pi} \frac{e^{e^{i \phi}}}{5 - 4\cos(\theta - \phi)} d\phi =\\
\int_{0}^{2\pi} \frac{e^{e^{i \phi}}}{5 - 4\frac{e^{i(\theta - \phi)}+e^{-i(\theta - \phi)}}{2}} d\phi=
\int_{0}^{2\pi} \frac{e^{e^{i \phi}}}{5 - 2\left(e^{i(\theta - \phi)}+e^{-i(\theta - \phi)}\right)} d\phi=\\
-i \int_{0}^{2\pi} \frac{e^{e^{i \phi}}}{e^{i \phi}\left(5 - 2\left(e^{i(\theta - \phi)}+e^{-i(\theta - \phi)}\right)\right)} d\left(e^{i \phi}\right)=\\
-i \int_{0}^{2\pi} \frac{e^{e^{i \phi}}}{5e^{i \phi} - 2\left(e^{i\theta}+e^{i 2\phi}e^{-i\theta}\right)} d\left(e^{i \phi}\right)=\\
-i \int\limits_{|z|=1} \frac{e^{z}}{5z - 2\left(e^{i\theta}+z^2e^{-i\theta}\right)} dz=
i \int\limits_{|z|=1} \frac{e^{z}}{e^{-i \theta} \left(z-2 e^{i\theta}\right) \left(2z-e^{i \theta}\right)} dz=\\
i 2\pi i \operatorname {Res} \left(\frac{e^{z}}{e^{-i \theta} \left(z-2 e^{i\theta}\right) \left(2z-e^{i \theta}\right)},\frac{e^{i \theta}}{2}\right)=\\
-2\pi \operatorname {Res} \left(\frac{e^{z}}{e^{-i \theta} \left(z-2 e^{i\theta}\right) \left(2z-e^{i \theta}\right)},\frac{e^{i \theta}}{2}\right)=\\
-2\pi \cdot \lim\limits_{z\rightarrow \frac{e^{i \theta}}{2}} \left(z-\frac{e^{i \theta}}{2}\right)\cdot \frac{e^{z}}{e^{-i \theta} \left(z-2 e^{i\theta}\right) \left(2z-e^{i \theta}\right)}=\\
-\pi \cdot \lim\limits_{z\rightarrow \frac{e^{i \theta}}{2}} \frac{e^{z}}{e^{-i \theta} \left(z-2 e^{i\theta}\right)}=
-\pi \frac{e^{\frac{e^{i \theta}}{2}}}{e^{-i \theta} \left(\frac{e^{i \theta}}{2}-2 e^{i\theta}\right)}=\frac{2\pi}{3} e^{\frac{e^{i \theta}}{2}}$$

Also, I think there is a typo in the original question, let's say $\theta=1$ we have 


*

*wolfram result here $\frac{2\pi}{3} e^{\frac{e^{i\cdot 1}}{2}} = \color{red}{2.504...} + i 1.12... $

*wolfram result here $\int\limits_{0}^{2\pi} \frac{e^{\cos(\phi)} \cos(\sin \phi)}{5 - 4\cos(1 - \phi)} d\phi=\color{red}{2.504...}$
A: Another method, which uses a Fourier series and avoids integration in the complex plane. As suggested, the problem is equivalent to calculate
\begin{equation}
 I=\int_{0}^{2\pi} \frac{e^{\exp(i\phi)}}{5 - 4\cos(\theta - \phi)} d\phi 
\end{equation} 
Changing  $x=\phi-\theta$ and using the $2\pi$-periodicity of the integrand,
\begin{equation}
 I=\int_{0}^{2\pi} \frac{e^{\exp(i\phi)\exp(ix)}}{5 - 4\cos x} dx
\end{equation} 
From the Fourier series
\begin{equation}
 \sum_{n=1}^{\infty} a^{n} \cos(nx) = \frac12\left(\frac{1-a^{2}}{1-2 a \cos x + a^{2}}-1\right)
\end{equation} 
using $a=1/2$, and denoting $z=e^{i\phi}$,
\begin{equation}
 I=\frac{1}{3}\int_{0}^{2\pi} e^{z\exp(ix)}\,dx+\frac{2}{3}\sum_{n=1}^{\infty}2^{-n}\int_{0}^{2\pi} e^{z\exp(ix)}\cos nx\,dx 
\end{equation} 
Then we have to evaluate integrals 
\begin{equation}
 I_k=\int_{0}^{2\pi} e^{z\exp(ix)}\cos kx\,dx
\end{equation} 
where $k$ is an integer. One may expand the exponential term :
\begin{equation}
 I_k=\sum_{p=0}^\infty\int_{0}^{2\pi} \frac{z^p\exp(ipx)}{p!}\cos kx\,dx
\end{equation} 
The only non vanishing terms in the $p$-summation correspond to $p=k$ for orthogonality reasons:
\begin{align}
 I_k&=\frac{\pi}{k!}z^k\text{ for } k\ne 0\\
 I_0&=2\pi
\end{align}
We obtain therefore
\begin{align}
 I&=\frac{2\pi}{3}+\frac{2\pi}{3}\sum_{n=1}^{\infty}\frac{2^{-n}z^n}{n!}\\
 &=\frac{2\pi}{3}e^{\exp(i\theta)/2}
\end{align} 
which is the result given in the answer by @rtybase 
