 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/4/d/ii/solution

Use

z = e^{i θ}

, so

sin θ = (z - z^{- 1}) / (2 i)

and

d θ = d z / (i z)

. The unit circle is traversed anticlockwise, and the integral becomes

\oint_{∣ z ∣ = 1} \frac{2 d z}{- 3 z ^{2} + 10 i z + 3} = \oint_{∣ z ∣ = 1} \frac{2 d z}{- 3 ( z - i /3 ) ( z - 3 i )} .

z = i /3

lies inside the contour. Its residue is

\frac{2}{- 6 ( i /3 ) + 10 i} = \frac{1}{4 i} .

Hence the residue theorem gives

\int_{0}^{2 π} \frac{d θ}{5 - 3 sin θ} = 2 πi \frac{1}{4 i} = \frac{π}{2} .

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