Integral using even function $$\int_{-{2 π}/3}^{{2 π/3}}-\frac {4\cos(x)}{1 + e^{-4x}} \,\mathrm{d}x $$
I have tried writing this as an even function so I can evaluate:
$$\int_0^{{2π}/3} f_\text{even}(x)\mathrm{d}x$$
But I'm struggling to get to this point as I end up with worse functions that I started with. Any help would be appreciated.
 A: Any function $f(x)$ can be decomposed as an even function $g(x) = g(-x)$ and an odd function $h(-x) = -h(x)$ as follows: 
\begin{equation}
g(x) = \frac{f(x) + f(-x)}{2}
\end{equation}
and 
\begin{equation}
h(x) = \frac{f(x) - f(-x)}{2},
\end{equation}
We can see that 
$f(x) = g(x) + h(x)$ and $g(-x) = g(x)$ and $h(-x) = -h(x)$ as desired. 
Then $\int_{-a}^a f(x) dx = \int_{-a}^a g(x)dx  + \int_{-a}^a h(x)  dx = 2 \int_{0}^a g(x) dx$. 
In your case we have 
\begin{equation}f(x) = -\frac{4\cos(x)}{1+\exp(-4x)}\end{equation}
and 
\begin{equation}
f(-x) = -\frac{4\cos(x)}{1+\exp(4x)}
\end{equation}
so by our previous definition 
\begin{align*}
g(x) &=-2\cos(x)(\frac{1}{1+\exp(-4x)} + \frac{1}{1+\exp(4x)}) \\&= -2\cos(x)(\frac{(1+\exp(4x)) + (1+\exp(-4x))}{(1+\exp(4x)) (1+\exp(-4x))}) \\&= -2\cos(x),
\end{align*}
which can be seen by expanding the denominator. Then your integral becomes: 
\begin{equation}
I = 2\int_0^{2\pi/3} dx ( -2 \cos(x)) = -4 (\sin(x))|_0^{2\pi/3} = -4\sin(2\pi/3) = -2\sqrt{3}
\end{equation}
A: Note that,
$$\frac {1}{1 + e^{-4x}} = \frac12+ \frac12 \tanh(2x)$$
Then,
$$\int_{-{2 π}/3}^{{2 π/3}}-\frac {4\cos x}{1 + e^{-4x}} dx 
= -4\int_0^{{2 π/3}}\cos x  dx $$
A: Your integrand is not an odd function. However, if you define
$$
g(x):=\frac12 - \frac1{1+e^{-4x}}=\frac{e^{-4x}-1}{2(e^{-4x}+1)},
$$
you can check that $g(-x)=-g(x)$, i.e. $g$ is an odd function. So your integral now equals
$$\int_{-2\pi/3}^{2\pi/3} 4\cos(x)\left[ g(x)-\frac12\right]\,dx,$$
which splits into the integral of an odd function plus the integral of an even function.
