Alternative ways to prove that $\int_0^{2\pi} \cos(\sin{x})e^{\cos{x}} dx=2\pi$ without complex analysis? $$\int_0^{2\pi} \cos(\sin{x})e^{\cos{x}} dx=2\pi$$
I derived this rather incredible result via Cauchy's theorem as I was working through some simple contour integrals. I was wondering if this integral can be solved without complex analysis, and if so how? I can't see any obvious substitutions that can be possible, nor any parametrisations that could simplify the integral through Feynman's method.

Notice first that the integrand is symmetrical about the axis $\theta=\pi$, so we have
$$\int_0^{2\pi}e^{\cos\theta}\cos(\sin\theta)d\theta = 2\int_0^{\pi}e^{\cos\theta}\cos(\sin\theta)d\theta$$
We define the following function:
\begin{align}
I(t) :&= \int_{0}^{\pi}e^{t\cos\theta}\cos(t\sin\theta)d\theta\\
&= \frac{1}{2}\int_{0}^{\pi}\exp\left[\frac{t}{2}(e^{i\theta}+e^{-i\theta})\right]\left[\exp\left(\frac{t}{2}(e^{i\theta}-e^{-i\theta})\right)+\exp\left(-\frac{t}{2}(e^{i\theta}-e^{-i\theta})\right)\right]d\theta\\
&= \frac{1}{2}\int_{0}^{\pi}\left[\exp\left(te^{i\theta}\right)+\exp\left(te^{-i\theta}\right)\right]d\theta
\end{align}
We differentiate with respect to $t$ and obtain
\begin{align}
I'(t) = \frac{1}{2}\int_0^\pi\left[e^{i\theta}\exp\left(te^{i\theta}\right) + e^{-i\theta}\exp\left(te^{-i\theta}\right)\right]d\theta
\end{align}
Let $u:=e^{i\theta}$, then $-idu=e^{i\theta}d\theta$, $u(0) = 1$ and $u(\pi)=-1$. Similarly, let $z:=e^{-i\theta}$, then $idz=e^{-i\theta}d\theta$, $z(0) = 1$ and $z(\pi)=-1$.
Then
\begin{align}
I'(t) &= -\frac{i}{2}\int_1^{-1}e^{tu}du + \frac{i}{2}\int_1^{-1}e^{tz}dz\\
&= \frac{i}{2}\left[\int_{-1}^1 e^{tu}du - \int_{-1}^1 e^{tz}dz\right] \\
&= 0
\end{align}
since $u$ and $z$ are dummy variables.
Since $I'(t) = 0$ for any $t$, $I(t) = c$ for some $c$ for all values of $t$.
We easily compute
$$I(0) = \int_0^\pi d\theta = \pi$$
Thus we have the following result:
$$\int_0^{2\pi}e^{\cos\theta}\cos(\sin\theta)d\theta = 2\pi$$
