I came across this integral in a text $$\int_{-1}^{1} \frac{cosx}{e^{\frac{1}{x}}+1}.$$
This was the approach used in the text:
Let $$g(x)= \int_{-1}^{1} \frac{\cos x}{e^{\frac{1}{x}}+1}$$
Then $g(x)$ is equivalent to $$g_{e}(x) + g_{o}(x),$$ that is, it is the sum of an even function $g_e$, and odd function $g_o$. Then $$\int_{-1}^{1}g(x) = \int_{-1}^{1}g_{e}(x) \ + \int_{-1}^{1}g_{o}(x)$$ Then it took $$\int_{-1}^{1}g_{o}(x) \text{ as } 0$$ Talked about symmetry of odd function from the point of view of integrals as area under a graph
Can someone please explain this better—I don't really get it. is this as a result of the limit of the integral $(1,-1)$ or is there any other reason?
If $f$ is an odd function then for all $a \geq 0$, $$I = \int_{-a}^a f(x) \mathrm{dx} = 0$$
Indeed, you can make the substitution $t=-x$ to get $$I = \int_{-a}^a f(x) \mathrm{dx} = \int_{a}^{-a} -f(-t) \mathrm{dt} = -\int_{-a}^a f(t) \mathrm{dt} = -I$$
So $I=-I$, so $I=0$.