 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-2/12/d/solution

Subtract the integrands first, using

cos u - cos^{3} u = cos u sin^{2} u

. Then set

s = sin u

I_{n, 1} (π /2, 0) - I_{n, 3} (π /2, 0) = \int_{0}^{1} s^{2} (1 - s^{3})^{n} d s .

w = s^{3}

makes

d w = 3 s^{2} d s

, so

I_{n, 1} (π /2, 0) - I_{n, 3} (π /2, 0) = \frac{1}{3} \int_{0}^{1} (1 - w)^{n} d w = \frac{1}{3 ( n + 1 )} .

This also holds at

n = 0

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