Today we consider the definite integral

$$ I=\int_{-\pi/2}^{\pi/2}\frac{e^{\vert\sin(x)\vert}\cos(x)}{1+e^{\tan(x)}}dx . $$

First Observation: The domain of integration is symmetric about the y- axis.

Second Observation: The denominator

$$ e^{\vert\sin(x)\vert}\cos(x) $$

is even.

So, a keen mathematician might try the reflection in coordinates

$$ x\rightarrow -t $$

so that the integral domain is unchanged, and the denominator of the integrand is unchanged.

Therefore, we have

$$ I=\int_{-\pi/2}^{\pi/2}\frac{e^{\vert\sin(t)\vert}\cos(t)}{1+e^{-\tan(t)}}dt . $$

Third Observation: It is a quite common and nice equality (it is simple but great!)

$$ \frac{1}{1+u}+\frac{1}{1+u^{-1}}=1 $$

for any u. Therefore, we have, by summing the integral twice,

$$ 2I=\int_{-\pi/2}^{\pi/2}(e^{\vert\sin(x)\vert}\cos(x))(\frac{1}{1+e^{\tan(x)}}+\frac{1}{1+e^{-\tan(x)}})dx \\= \int_{-\pi/2}^{\pi/2}e^{\vert\sin(x)\vert}\cos(x)dx = 2\int_0^{\pi/2}e^{\vert\sin(x)\vert}\cos(x)dx $$

, where the last equality holds because of observation 1. Therefore, we have

$$ I=\int_0^{\pi/2}e^{\sin(x)}\cos(x)dx=\int_0^1e^\eta d\eta=e-1 $$

where in the second equality, we have introduced a change in coordinates

$$ \sin(x)\rightarrow\eta $$