TITLE: How to prove that $\int\limits_{-\pi/2}^{\pi/2}\frac{\cos(2x)}{e^x+1}=0$?
QUESTION [7 upvotes]: I am stuck trying to show that $\displaystyle \int\limits_{-\pi/2}^{\pi/2}\frac{\cos(2x)}{e^x+1}=0$ 
I have tried using a Squeeze Theorem type approach, but at $\pi/4$ any function I choose overlaps and becomes less than or greater than the function depending. I am not sure where to start anymore. Anything will be helpful. Thank you in advance!

REPLY [5 votes]: There is a result 
stated as follows

Suppose $f$ is continuous and even on $[-a,a]$, $a>0$ then 
  $$\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$$

Using this result, $\displaystyle \int_{-\pi/2}^{\pi/2}\frac{\cos(2x)}{e^x+1}dx=\int_{0}^{\pi/2}\cos(2x)\,dx=0$.