Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2016 / S1 / 3

Set $z=e^{i\theta }$. The integral becomesThe roots are $z=(-3\pm 2)/(2-i)$, only $z_{+}=-1/(2-i)$ lying inside. The derivative of the quadratic at $z_{+}$ is two, so the residue is $1/(2i)$. Consequently $\boxed{{\int }_0^{2\pi }d\theta /(3+2\cos \theta +\sin \theta )=\pi }$. The denominator is positive since $3>\sqrt{5}$, so no principal-value qualification is required.

Cauchy's theorem says that a holomorphic function integrates to zero around a closed contour homologous to zero in its domain. Thus deforming one contour into another without crossing a singularity preserves the integral. More generally for a meromorphic $f$, the exact condition isEqual winding numbers about every singularity are sufficient. They are not necessary for one particular function because residues can cancel; even arbitrary contours have equal integrals for $f=0$. A topological deformation criterion must not be confused with a necessary condition for accidental equality.

The finite singularities are square-root branch points at $z=\pm 1$. Choose the cut between them and the branch $f(z)\sim z$ at infinity. The two banks then have values $+i\sqrt{1-x^2}$ above and $-i\sqrt{1-x^2}$ below. A counterclockwise loop around the cut traverses its upper bank right-to-left, so its integral is $-2iI$, where $I$ is the required real integral. At large radius,Therefore $\boxed{I=\pi /2}$. Infinity is not another square-root branch point: this branch has an ordinary pole there on its sheet. Small detours about the finite endpoints vanish as their radius to power $3/2$.