I read this question. The integral has a special property, so it might possibly be evaluable? No one tried evaluating it, so I created this. Not very often I ask question like this, but here it is.
So what is,
$$\int_0^{\pi} \frac{e^{\sin x}\cos(x)}{1+e^{\tan x}} \, dx$$
A list of Naïve Ideas
Sub $u=\sin(x)$ allows you to cancel the $\cos$.
Half angle substitution
List the obscure special function it equals.
Try contour integration ;) (just being stupid here)
(Disclaimer: I have no reason to believe this integral has closed form beyond the property mentioned in the link)
Extending a comment made earlier:
As the integral was first defined, with $\phi(x) = e^{|\sin x|} \, \cos x$, an identity was presented as, in corrected form, \begin{align} \int_{0}^{\pi} \frac{\phi(x) \, dx}{1 + e^{\tan x}} = - \frac{1}{2} \, \int_{0}^{\pi} \tanh\left(\frac{\tan x}{2} \right) \, \phi(x) \, dx. \end{align} Wolfram Alpha provides a numerical value of $- .727486$ for the integral on the left and \begin{align} \int_{0}^{\pi} \tanh\left(\frac{\tan x}{2} \right) \, \phi(x) \, dx = 1.45497. \end{align} In either case the numerical values lead to the speculation that \begin{align} \int_{0}^{\pi} \frac{\phi(x) \, dx}{1 + e^{\tan x}} = - \frac{1}{2} \, \int_{0}^{\pi} \tanh\left(\frac{\tan x}{2} \right) \, \phi(x) \, dx \approx - \frac{672}{855}. \end{align} Further work can be done to evaluate the integral.