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

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

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.

ii

Solution

The difference and sum of the two antiderivatives in part (i) give

\int e^{x} (sin x - cos x) d x = - e^{x} cos x + C,

and

\int e^{x} (sin x + cos x) d x = e^{x} sin x + C .

Their derivatives also verify the signs immediately.

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