Integral with Composite Trigonometric Function 
Find the value of $$I=\int_0^{\large\pi} e^{\cos\theta}\cos(\sin\theta)\ d\theta$$


My Attempts:
Using $$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)\ dx $$ We have
$$I=\int_{0}^{\large \pi}e^{-\cos\theta}\,\cos(\sin\theta)\ d\theta$$
Adding the two integrals we get
$$I=\int_{0}^{\large\pi}\cosh(\cos\theta)\cos(\sin\theta)\ d\theta$$ $\implies$
$$I=2\int_{0}^{\Large\frac{\pi}{2}}\cosh(\cos\theta)\cos(\sin\theta)\ d\theta$$ Please help me from here.
 A: Rewrite
$$
\int_0^{\large\pi} e^{\cos\theta}\cos(\sin\theta)\ d\theta=\Re\left[\int_0^{\large\pi} e^{\Large e^{i\theta}}d\theta\right].
$$
Let
$$
I(\alpha)=\int_0^{\large\pi} e^{\Large\alpha e^{i\theta}}d\theta,
$$
then
$$
\frac{dI}{d\alpha}=I'(\alpha)=\int_0^{\large\pi} e^{i\theta}e^{\Large\alpha e^{i\theta}}d\theta.
$$
Rewrite
$$
I'(\alpha)=\frac{1}{i\alpha}\int_0^{\large\pi} i\alpha e^{i\theta}e^{\Large\alpha e^{i\theta}}d\theta.
$$
Let $x=\alpha e^{i\theta}\;\color{blue}{\Rightarrow}\;dx=i\alpha e^{i\theta}\ d\theta$, then
$$
I'(\alpha)=\frac{1}{i\alpha}\left[e^{\Large\alpha e^{i\theta}}\right]_0^{\large\pi}=\frac{e^{-\large\alpha}-e^{\large\alpha}}{i\alpha}=\frac{2\sinh\alpha}{\alpha}i
$$
Thus $\Re\left[I'(\alpha)\right]=0$ and $I(\alpha)$ is a constant.
Taking $\alpha=0$ yields $I(0)=\pi$. Consequently
$$
\int_0^{\large\pi} e^{\cos\theta}\cos(\sin\theta)\ d\theta=\large\color{blue}{\pi}.
$$
A: It is just $\pi$. You can write your integral as:
$$ I =\Re\int_{0}^{\pi}\exp\left(\exp(i\theta)\right)\,d\theta, $$
then, using the Taylor series of the exponential function:
$$ I = \Re\sum_{k=0}^{+\infty}\frac{1}{k!}\int_{0}^{\pi}\exp(ki \theta)\,d\theta, $$
where all the terms corresponding to even values of $k$ (except the very first one, $k=0$) are zero and all the terms corresponding to odd values of $k$ are pure imaginary. This simply gives:
$$ I=\pi.$$
