 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2/18/a/solution

Write

I_{n} = \int sin^{n} x d x

sin^{n - 1} x

and

sin x d x

I_{n} = - cos x sin^{n - 1} x + (n - 1) \int sin^{n - 2} x cos^{2} x d x .

Using

cos^{2} x = 1 - sin^{2} x

on the right gives

I_{n} = - cos x sin^{n - 1} x + (n - 1) (I_{n - 2} - I_{n})

. Rearranging proves

I_{n} = - \frac{cos x sin ^{n - 1} x}{n} + \frac{n - 1}{n} I_{n - 2} + C .

For

J_{n} = \int cos^{n} x d x

cos^{n - 1} x

and

cos x d x

J_{n} = sin x cos^{n - 1} x + (n - 1) (J_{n - 2} - J_{n})

. Hence

J_{n} = \frac{sin x cos ^{n - 1} x}{n} + \frac{n - 1}{n} J_{n - 2} + C .

An arbitrary constant of integration absorbs the choices in the lower-order antiderivatives.

Starting from

I_{0} = J_{0} = x

, successive use of

n = 2, 4, 6

gives

\int sin^{6} x d x = \frac{5 x}{16} - \frac{cos x sin ^{5} x}{6} - \frac{5 cos x sin ^{3} x}{24} - \frac{5 sin x cos x}{16} + C,

\int cos^{6} x d x = \frac{5 x}{16} + \frac{sin x cos ^{5} x}{6} + \frac{5 sin x cos ^{3} x}{24} + \frac{5 sin x cos x}{16} + C .

These expressions follow directly from the requested recurrence, rather than requiring a separate trigonometric expansion.

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