# contour integration of trigonometric function limit is not pi or 2pi

I am using contour integration to find $$\int_0^{2p} \frac{1}{2+cos\theta}\,d\theta~.$$

I learned in the book that upper limit of integral must be $$2\pi$$. To convert it to $$2\pi$$ i use substitution $$t=(\pi/p)\theta$$ but this change to $$(p/\pi)\int_0^{2\pi} \frac{1}{2+cos(pt/\pi)}\,dt~.$$

Now, here I can use contour integration by using $$e^{ipt/\pi} = z$$ and hence $$cos(pt/\pi) = 1/2z(z^2+1)$$ and the equation changes to $$1/i \int_0^{2\pi}\frac{2z}{z^2+4z+1}\,dz$$

Now this result has value which is not dependent on $$p$$. Am I doing something wrong, as the final result should have $$p$$. Please suggest.

Let's assume that $$0< p<\pi$$. Then, upon making the substitution $$z=e^{i\theta}$$, we have
$$\int_0^{2p} \frac1{2+\cos(\theta)}\,d\theta = -i\int_1^{e^{i2p}}\frac{2}{z^2+4z+1}\,\,dz\tag1$$
Note that the integral on the right-hand side of $$(1)$$ is not a closed contour and we cannot use the residue theorem to evaluate it.