Does an analytical solution exist for the value of integral? Be puzzling with this integral and wondering if an analytical solution exists. If it does, it has escaped me to date!
$$\int_{0}^{2\pi} \frac{1}{\exp(\sin(x))+\exp(\cos(x))}\, dx.$$ 
 A: This is not an answer.
The little scary problem gives (numerically)
$$\int_{0}^{2\pi} \frac{dx}{\exp(\sin(x))+\exp(\cos(x))}=3.183807032899083924101596594681\cdots$$
which (thanks to inverse symbolic calculators) is "close" to
$$\frac{ -20+10 e+45 e^2-59 \sqrt{1+e}-23 \sqrt{1+e^2}+10 \pi -20 \pi ^2+2 \sqrt{1+\pi }+23
   \sqrt{1+\pi ^2}} {23 }$$ the relative error being $2.43\times 10^{-19}$ percent.
A: $$I=\int_0^{2\pi}\frac{1}{\exp(\sin(x))+\exp(\cos(x))}dx$$
$$I(a,b)=\int_0^{2\pi}\frac{1}{a\exp(\sin(x))+b\exp(\cos(x))}dx$$

$$\frac{dI(a,b)}{da}=\int_0^{2\pi}\frac{-\exp(\sin(x))}{\left[a\exp(\sin(x))+b\exp(\cos(x))\right]^2}dx$$
$$\frac{dI(a,b)}{db}=\int_0^{2\pi}\frac{-\exp(\cos(x))}{\left[a\exp(\sin(x))+b\exp(\cos(x))\right]^2}dx$$

$$-(I_a+I_b)=\int_0^{2\pi}\frac{\exp(\sin(x))+\exp(\cos(x))}{\left[a\exp(\sin(x))+b\exp(\cos(x))\right]^2}dx$$
so at $(a,b)=(1,1)$ then $-(I_a+I_b)=I$. Maybe this can be manipulated and solved as a differential equation?

Also notice that using the well known rule:
$$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$
we get:
$$I=\int_0^{2\pi}\frac{1}{\exp(-\sin(x))+\exp(\cos(x))}dx$$
