why is $\int_{\pi/2}^{5\pi/2}\frac{e^{\arctan(\sin x)}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}}=\pi$?

Question

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.

Answer 1

Let the integrand be denoted by $f(x)$, and let $I$ be the value of this integral.

1. The integrand is $2\pi$-periodic so, $I=\int_T^{T+2\pi}f(x)dx$ for every $T$.
2. $f(x)+f(\frac{\pi}{2}-x)=1$.
3. Thus $2I=\int_0^{2\pi}dx=2\pi$.

Answer 2

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: $$\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$$