The calculation does not use complex variable function theory $\int _0^{2\pi }e^{\cos \left(x\right)}\cos \left(\sin \left(x\right)\right)dx$ Problem: calculate $\int _0^{2\pi }e^{\cos \left(x\right)}\cos \left(\sin \left(x\right)\right)dx$
This is a problem which post in How to evaluate $\int_{0}^{2\pi}e^{\cos \theta}\cos( \sin \theta) d\theta$?
I wish to find a calculation that does not use complex function theory. I want to prove the most elementary way. 
Please watch it for me. I sincerely thank you.
 A: Feynman’s Integration Technique without Contour Integration
First of all, we rewrite the integral as
\begin{aligned}
I & =\int_0^{2 \pi} e^{\cos x} \cos (\sin x) d x =\operatorname{Re} \int_0^{2 \pi} e^{\cos x+i \sin x} d x = \operatorname{Re} \int_0^{2 \pi} e^{e^{x i}} d x
\end{aligned}
Now we define a parametrized integral
$$I(a)=\int_0^{2 \pi} e^{a e^{x i}} d x$$
As usual, we differentiate $I(a)$ w.r.t. $a$, then perform integration by parts and get
\begin{aligned}
I^{\prime}(a) & =\int_0^{2 \pi} a i e^{x i} e^{a e^{x i}} d x \\
& =\int_0^{2 \pi} e^{a e^{x i}} d\left(e^{a e^{x i}}\right) \\
& =\left[e^{a e^{x i}}\right]_0^{2 \pi} \\
& =e^a-e^a \\
& =0
\end{aligned}
So we have for any $a$, $$I(a)=C$$ where $C$ is a constant.
Hence $$\int_0^{2 \pi} e^{\cos x} \cos (\sin x) d x  =I(1)=I(0)=2\pi $$
In general,
$$\boxed{\int_0^{2 \pi} e^{n\cos x} \cos (n\sin x) d x  =I(n)=I(0)=2\pi}$$
Wish it helps!
A: Via Fourier expansion the integrand equals to $\sum _{n=0}^{\infty } \frac{\cos (n x)}{n!}$ According to orthogonality of trig functions one has $I=\frac{\int_0^{2 \pi } \cos (0 x) \, dx}{0!}=2\pi$.
