 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-1/14/h/solution

Square the two-term form and integrate directly:

\int_{0}^{2 π} (sin x - 3 cos x)^{2} d x = \int_{0}^{2 π} sin^{2} x d x + 3 \int_{0}^{2 π} cos^{2} x d x - 23 \int_{0}^{2 π} sin x cos x d x .

The first two integrals are each

π

before their prefactors, while the cross term is zero because its integrand is

\frac{1}{2} sin 2 x

. Thus the result is again $4 π$ , agreeing with Parseval's identity.

 New to topics? Read the docs here!