Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S1 / 2 / b / Solution

Put $z=e^{i\theta }$, so $\sin \theta =(z-z^{-1})/(2i)$ and $d\theta =dz/(iz)$. The integral becomesThe poles are $-2i/3$ and $-3i/2$; only the first is inside the unit circle. The derivative of the denominator there is $12(-2i/3)+13i=5i$. The residue theorem therefore givesThere are no poles on the contour because the real denominator is at least one.