Split integral with $e^{\sin t}$ From this answer:
$$\int_{0}^{2\pi} e^{\sin t} = \int_{0}^{\pi} e^{\sin t}\ dt+\int_{0}^{\pi} e^{-\sin t}\ dt$$
How to prove this?

My attempt:
$e^{-\sin t}$ is neither odd, nor even; the usual way to split such an integral could be
$$\int_{0}^{2\pi} e^{\sin t} = \int_{0}^{\pi} e^{\sin t}\ dt+\int_{\pi}^{2\pi} e^{\sin t}\ dt$$
The second addend could become, putting $u = -t$, $du = -dt$ and $\sin(-t) = -\sin(t)$:
$$- \int_{-2\pi}^{-\pi} e^{-\sin t}\ dt$$
but however it is not like in the initial expression.

You have that $\sin(t)=-\sin (t-\pi)$. Apply to the integral over  $(\pi,2\pi)$ and change variables $s=t-\pi$. 
