 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/a/i/solution

Write

I = \int e^{x} sin x d x

and

J = \int e^{x} cos x d x

. Two applications of integration by parts, first taking the sine and then the cosine as the differentiated factor, give

I = e^{x} sin x - J + C_{1}, J = e^{x} cos x + I + C_{2} .

Thus

I + J = e^{x} sin x

and

J - I = e^{x} cos x

, in each case up to an arbitrary constant of integration. Keeping that freedom avoids identifying independently chosen antiderivatives too rigidly.

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