 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/b/solution

s = sin u

, so

d s = cos u d u

, and write

J_{n} = I_{n, 1} (π /2, 0)

J_{n} = \int_{0}^{1} (1 - s^{3})^{n} d s

. For

n > 0

J_{n} = [s (1 - s^{3})^{n}]_{0}^{1} + 3 n \int_{0}^{1} s^{3} (1 - s^{3})^{n - 1} d s .

Both endpoint terms vanish. Use

s^{3} = 1 - (1 - s^{3})

inside the remaining integral:

J_{n} = 3 n (J_{n - 1} - J_{n}) .

Consequently

I_{n, 1} (π /2, 0) = \frac{3 n}{3 n + 1} I_{n - 1, 1} (π /2, 0), f (n) = \frac{3 n}{3 n + 1} .

This is the

p = 3

instance of the power-complement integral recurrence, with initial value

J_{0} = 1

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