Proof of an integral involving even and odd functions

This note provides a proof of the following identity

\[\int_{-a}^a \frac {f(x)}{1+c^{g(x)}} \ dx = \int_0^a f(x)\ dx\]

where $f(x)$ is an even function, $g(x)$, an odd function, and $a$ and $c$ are constants.

Proof:

$\begin{aligned} \int_{-a}^a \frac {f(x)}{1+c^{g(x)}} \ dx & = {\color{#3D99F6} \int_{-a}^0 \frac {f(x)}{1+c^{g(x)}} \ dx} + \int_0^a \frac {f(x)}{1+c^{g(x)}} \ dx & \small \color{#3D99F6} \text{Let }u=-x \implies dx = -du \\ & = {\color{#3D99F6} - \int_a^0 \frac {f(-u)}{1+c^{g(-u)}} \ du} + \int_0^a \frac {f(x)}{1+c^{g(x)}} \ dx & \small \color{#3D99F6} \text{Since for even function }f(-u) = f(u) \\ & = {\color{#3D99F6} \int_0^a \frac {f(u)}{1+c^{-g(u)}} \ du} + \int_0^a \frac {f(x)}{1+c^{g(x)}} \ dx & \small \color{#3D99F6} \text{ and odd function }g(-u) = - g(u) \\ & = {\color{#3D99F6} \int_0^a \frac {c^{g(u)} f(u)}{1+c^{g(u)}} \ du} + \int_0^a \frac {f(x)}{1+c^{g(x)}} \ dx & \small \color{#3D99F6} \text{Replace }u \text{ with }x \\ & = \int_0^a \frac {\left(1+c^{g(x)}\right) f(x)}{1+c^{g(x)}} \ dx \\ & = \int_0^a f(x) \ dx \quad \blacksquare \end{aligned}$