Definite integral of an odd function is 0 (symmetric interval)
Because $f$ is odd, $f(-x) = -f(x)$. Thus, we can write
$$ \int_{x=-a}^0 f(x) \, dx = \int_{x=-a}^0 -f(-x) \, dx $$
Let $u = -x, du = -dx$, then
$$ \int_{x=-a}^0 -f(-x) \, dx = \int_{u=a}^0 f(u) \, du $$
Reversing the limits of integration inverts the result, so
$$ \int_{u=a}^0 f(u) \, du = -\int_{u=0}^a f(u) \, du $$
as desired. (The $u$ is just a dummy variable; it can be replaced by $x$ at this point, despite our originally having assigned it as $u = -x$.)