 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/17/b/solution

Let

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

and

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

. Integrating the trigonometric functions and differentiating

e^{n x}

I_{n} = - e^{n x} cos x + n J_{n}, J_{n} = e^{n x} sin x - n I_{n} .

Substitution yields

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

and

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

. Therefore

\int e^{n x} (sin x - cos x) d x = I_{n} - J_{n} = - \frac{e ^{n x}}{n ^{2} + 1} [(1 + n) cos x + (1 - n) sin x] + C .

n

was used, so this formula also applies at

n = 0

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