TITLE: Evaluating a definite integral involving $ \tan^{-1}$
QUESTION [6 upvotes]: The question is to evaluate $$\int_{\pi/2}^{5\pi/2} \frac{e^{\tan^{-1} \sin x}}{e^{\tan^{-1} \sin x}+e^{\tan^{-1} \cos x}}dx$$
I tried to take idea from the graph of $\tan^{-1} \tan x$ and rewrite the integral as $$\int_{\pi/2}^{5\pi/2} \frac{e^{\tan^{-1} \sin x}}{e^{\tan^{-1} \sin x}+e^{\tan^{-1} -\cos x}}dx$$.I couldn't proceed from here.Any ideas?Thanks.

REPLY [5 votes]: Upon enforcing the substitution $x=\pi/2 -y$ we see that 
$$\begin{align}
I&=\int_{\pi/2}^{5\pi/2}\frac{e^{\arctan(\sin(x))}}{e^{\arctan(\sin(x))}+e^{\arctan(\cos(x))}}\,dx\\\\
&=\int_0^{2\pi}\frac{e^{\arctan(\cos(y))}}{e^{\arctan(\cos(y))}+e^{\arctan(\sin(y))}}\,dy\tag 1
\end{align}$$
Noting that the integrand is $2\pi$-periodic, we can write 
$$\int_0^{2\pi}\frac{e^{\arctan(\cos(y))}}{e^{\arctan(\cos(y))}+e^{\arctan(\sin(y))}}\,dy=\int_{\pi/2}^{5\pi/2}\frac{e^{\arctan(\cos(y))}}{e^{\arctan(\cos(y))}+e^{\arctan(\sin(y))}}\,dy \tag 2$$
Using $(2)$ in $(1)$ we see that 
$$I=\int_{\pi/2}^{5\pi/2}\frac{e^{\arctan(\cos(x))}}{e^{\arctan(\sin(x))}+e^{\arctan(\cos(x))}}\,dx  \tag 3$$
Therefore, by adding $(1)$ and $(3)$ and dividing by $2$ yields the coveted result
$$\int_{\pi/2}^{5\pi/2}\frac{e^{\arctan(\sin(x))}}{e^{\arctan(\sin(x))}+e^{\arctan(\cos(x))}}\,dx=\pi$$