Integration of $\int_{-\pi/4}^{\pi/4} \frac{\sec^2x}{1+e^x}dx$

Question

Actually I got stuck in a integration it is $$\int_{-\pi/4}^{\pi/4} \frac{\sec^2x}{1+e^x}dx$$

I tried multipying and dividing by $\tan x$ and then substitution but nothing worked for me. I am not understanding the way to solve it. Please give me a hint to start for this integration.

Answer 1

Hint
For a even function $f(x)$, we have $$\int_{-a}^a\frac{f(x)}{1+e^x}dx\\ =\frac12\left(\int_{-a}^a\frac{f(x)}{1+e^x}dx+\int_{-a}^a\frac{f(x)}{1+e^x}dx\right)\\ =\frac12\left(\int_{-a}^a\frac{f(x)}{1+e^x}dx+\int_{-a}^a\frac{f(-x)}{1+e^{-x}}dx\right)\\ =\frac12\left(\int_{-a}^a\frac{f(x)}{1+e^x}dx+\int_{-a}^a\frac{f(x)}{1+e^{-x}}dx\right)\\ =\frac12\int_{-a}^a\left(\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}\right)f(x)dx\\ =\frac12\int_{-a}^a1\cdot f(x)dx$$ Now, substitute $a=\frac{\pi}4$ and $f(x)=\sec^2x$.