Evaluating $\int_0^\pi\frac{e^{-\sin x}}{e^{-\sin x}+e\sin x}dx$

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I think I thought of this integral a little over a year ago and I just haven't been able to do it. I really want to and everytime I sit down with it I learn a little bit more (or atleast I think I do) and then I get really stumped and give up.

The integral in question is

$$I=\int_{0}^{\pi}\frac{e^{-\sin(x)}}{e^{-\sin(x)}+e\sin(x)}dx$$

Which can be rewritten into

$$I=2\int_{0}^{\pi/2}\frac{e^{-\sin(x)}}{e^{-\sin(x)}+e\sin(x)}dx$$

by making note of the fact that the function enclosed in the integral is even.

One other thing to note is that if we consider the $esin(x)$ term in the denominator and write a "new" integral with this term replacing the $e^{-\sin(x)}$ term in the numerator we get the following

$$J=\int_{0}^{\pi}\frac{e\sin(x)}{e^{-\sin(x)}+e\sin(x)}dx.$$

Taking the sum of $I$ and $J$ we, of course, get the answer $\pi$, meaning that a solution to either one would give a solution to the other (hopefully this helps in some way). I have a decent background and level of understanding with many of the more popular "advanced" integration techniques and special functions, so don't be afraid to write a really scary looking solution. Finally, I genuinely don't think there is a nice closed-form solution for this integral, but I don't know how to show that and am really hoping there is a closed-form.

Best of luck with your solutions.