TITLE: why is $\int_{\pi/2}^{5\pi/2}\frac{e^{\arctan(\sin x)}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}}=\pi$?
QUESTION [5 upvotes]: I cannot make progress on the definite integral $$\int_{\pi/2}^{5\pi/2}\frac{e^{\arctan(\sin x)}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}}\,dx=\pi$$
I know the result is $\pi$ from numerical approximation. Could someone give some hints? Is there a clever substitution I'm missing? I'd prefer hints to a full solution.

REPLY [7 votes]: Another approach is to write:
$$\begin{aligned}  I & =  \int_{\pi/2}^{5\pi/2} \frac{e^{\arctan(\sin x)}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}}\;{dx} \\& =  \int_{0}^{5\pi/2} \frac{e^{\arctan(\sin x)}\;{dx}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}}-\int_{0}^{\pi/2} \frac{e^{\arctan(\sin x)}\;{dx}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}} \\& = I_{1}-I_{2} \end{aligned}$$ 
Both the integrals can be easily evaluated using the well known definite integral property which states that:
$$\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$$