This works because the even part of $\displaystyle{\frac{1}{1+e^x}}$ is $\frac{1}{2}$.

If $g:[-a,a]\to \mathbb R$ is a function, then $g$ has a unique representation as a sum of an even and an odd function, $g=h+k$, with $h(-x)=h(x)$ and $k(-x)=-k(x)$. If $f:[-a,a]\to\mathbb R$ is even, then $g(x)f(x)=h(x)f(x)+k(x)f(x)$ has even part $h(x)f(x)$ and odd part $k(x)f(x)$. Since the integral of an odd function on $[-a,a]$ is zero and the integral of an even function on $[-a,a]$ is twice the integral on $[0,a]$, this yields

$$\int_{-a}^a g(x)f(x)dx=\int_{-a}^ah(x)f(x)dx=2\int_0^a h(x)f(x)dx.$$

As has been seen in previous questions on this site (like this one) the formula for $h$ is $h(x)=\frac{1}{2}(g(x)+g(-x))$. In this problem, $\displaystyle{g(x)=\frac{1}{1+e^x}}$, and $h(x)=\frac{1}{2}$.

This works because the even part of $\displaystyle{\frac{1}{1+e^x}}$ is $\frac{1}{2}$.

If $g:[-a,a]\to \mathbb R$ is a function, then $g$ has a unique representation as a sum of an even and an odd function, $g=h+k$, with $h(-x)=h(x)$ and $k(-x)=-k(x)$. If $f:[-a,a]\to\mathbb R$ is even, then $g(x)f(x)=h(x)f(x)+k(x)f(x)$ has even part $h(x)f(x)$ and odd part $k(x)f(x)$. Since the integral of an odd function on $[-a,a]$ is zero and the integral of an even function on $[-a,a]$ is twice the integral on $[0,a]$, this yields

$$\int_{-a}^a g(x)f(x)dx=\int_{-a}^ah(x)f(x)dx=2\int_0^a h(x)f(x)dx.$$

As has been seen in previous questions on this site (like this one) the formula for $h$ is $h(x)=\frac{1}{2}(g(x)+g(-x))$. In this problem, $\displaystyle{g(x)=\frac{1}{1+e^x}}$, and $h(x)=\frac{1}{2}$.