The tangent half-angle substitution

Idea. The substitution $t=\tan\frac x2$ turns any rational expression in $\sin x$ and $\cos x$ into a rational function of $t$.

Substitution. For $x\in[0,\pi)$, $t$ ranges over $[0,+\infty)$ and
$$\frac1{2+\cos x}=\frac1{2+\frac{1-t^2}{1+t^2}}=\frac{1+t^2}{3+t^2},\qquad dx=\frac{2\,dt}{1+t^2}.$$

Computation.
$$\int_0^\pi\frac{dx}{2+\cos x}=\int_0^{+\infty}\frac{2\,dt}{3+t^2}=\frac2{\sqrt3}\left[\arctan\frac t{\sqrt3}\right]_0^{+\infty}=\frac2{\sqrt3}\cdot\frac\pi2=\frac\pi{\sqrt3}.$$

Check. The integrand lies between $\frac13$ and $1$, so the integral is between $\frac\pi3\approx1.05$ and $\pi$. The value $\frac\pi{\sqrt3}\approx1.81$ is plausible (the integrand is about $\frac12$ in the middle).