Integral using even function

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}

Note that,

$$\frac {1}{1 + e^{-4x}} = \frac12+ \frac12 \tanh(2x)$$

Then,

$$\int_{-{2 π}/3}^{{2 π/3}}-\frac {4\cos x}{1 + e^{-4x}} dx = -4\int_0^{{2 π/3}}\cos x dx $$

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.