I'm being asked to calculate $\int_{∂R}{\frac{Log(1+z)}{cosz-1}}dz$ where R is defined as being $R = \{ z ∈ C |− 1 / 2 ≤ \Re z ≤ 2 π + 1 / 2 , | \Im z| ≤ 1 |\}$.
Here I believe that I need to use Cauchy's residual formula to sum up my second degree residuals.
However from looking at similar exercises it seems to me that I only need to sum up the specific residuals that lie in $R$, something which I am not able to pinpoint.
Some help on another question led me to find that the second order residuals are equal to $\frac{-2}{1+2\pi}$
Precisely. You want to use the residue theorem so only the residues of points which lie inside the contour you're using (in this case a rectangle) will contribute to your integral.
The zeros of $cos(z)-1 = (e^{iz}+e^{-iz})/2 - 1$ are where you will calculate the residues, however they occur periodically (at $2\pi k)$, which is why the contour is restricted to a rectangle which only contains one such point, namely $2\pi$. Therefore your integral becomes:
$$\int_{\partial R} \frac{log(z+1)}{cos (z) -1} = 2\pi i \text{ Res}_{z = 2\pi}f(z)$$
Also it's not really relevant to you because of your chosen contour, but its important in problems with a complex log to make sure you can define a branch cut that doesn't interfere with your contour. The easisest branch cut of $log(1+z)$ would be $(-\infty,-1)$ so your contour is inside the simply connected domain $\mathbb{C} \setminus (-\infty,-1)$, and you can therefore proceed as if $log(1+z)$ is holomorphic.