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.

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    $\begingroup$ I have an idea is show that $$\int_{\pi/2}^{5\pi/2}\frac{e^{\arctan(\sin x)}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}}\,dx=\int_{\pi/2}^{5\pi/2}\frac{e^{\arctan(\cos x)}}{e^{\arctan(\sin x)}+e^{\arctan(\cos x)}}\,dx$$ $\endgroup$ – AsdrubalBeltran May 4 '14 at 19:13

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 $2I=\int_0^{2\pi}dx=2\pi$.
  • $\begingroup$ This is great +1 $\endgroup$ – Jeff Faraci May 4 '14 at 18:57

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


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