 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-2/16/a/solution

x = sin θ

, with

- π /6 \leq θ \leq π /6

. On this interval

cos θ > 0

, so it cancels the positive denominator square root. The transformed integral is

\frac{1}{2} \int_{- π /6}^{π /6} 1 - \frac{4}{9} sin^{2} θ d θ .

Its integrand is even. Therefore the answer in the PDF's modulus-amplitude convention for the elliptic integral of the second kind is

E (\frac{2}{3}, \frac{π}{6}) .

The modulus is

2/3

, not the squared parameter

4/9

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