Q:

Why does the odd part of the function integrate to $0$?

When evaluating the integral $$\int^1_{-1}\frac{\cos(x)}{1+e^{1/x}}dx,$$ the author says that the odd part of the function integrates to $0$ over $-1$ to $1$. I have calculated the even part of the function ($\frac{1}{2}\cos(x)$), so the integral is equal to $\sin(1)$.
But why exactly does the odd part integrate to $0$?

A:

If $I:=\int_0^ao(x)dx$ is finite with $o$ odd, $y=-x$ obtains $\int_{-a}^0o(y)dy=\int_0^ao(-x)dx=-\int_0^ao(x)dx=-I$, so $\int_{-a}^ao(x)dx=-I+I=0$.

