 Solution

ID: 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 θ}

, so

sin θ = (z - z^{- 1}) / (2 i)

and

d θ = d z / (i z)

. The integral becomes

\oint_{∣ z ∣ = 1} \frac{d z}{6 z ^{2} + 13 i z - 6} .

The poles are

- 2 i /3

and

- 3 i /2

; only the first is inside the unit circle. The derivative of the denominator there is

12 (- 2 i /3) + 13 i = 5 i

. The residue theorem therefore gives

\int_{0}^{2 π} \frac{d θ}{13 + 12 sin θ} = \frac{2 π}{5} .

There are no poles on the contour because the real denominator is at least one.

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