An integral that involves trigonometric substitutions (probably)

Question

So I have an integral: $\int _0 ^{2\pi} \frac{1}{3+cosx} dx$. I reckon it is good idea to use this substitution $t = tan\frac{x}{2}$ and so the integral becomes $\int \frac{1}{t^2 + 2}dt$. This integral should be okay, but I am confused with the new integration limits. When $x=2\pi$, $t=0$; when $x = 0$, $t=0$. This does not make sense. What is the problem with the limits?

Answer 1

Since$$\int_0^{2\pi}\frac1{3+\cos x}\,\mathrm dx=\int_{-\pi}^\pi\frac1{3+\cos(x+\pi)}\,\mathrm dx=2\int_0^\pi\frac1{3-\cos x}\,\mathrm dx$$(since $\cos$ is an even function), all you have to do is to compute the integral$$\int_0^\pi\frac1{3-\cos x}\,\mathrm dx,$$in which case your method works just fine.