Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 18 / a

Let

I_{n}

denote the integral with the cosine factor and

J_{n}

the one with the sine factor. In integration by parts, take

u = cos (n x)

and

d v = e^{x} d x

. Then

I_{n} = [e^{x} cos (n x)]_{0}^{π} + n J_{n} = (- 1)^{n} e^{π} - 1 + n J_{n} .

The integer restriction gives

cos (nπ) = (- 1)^{n}

ii

For

J_{n}

, take

u = sin (n x)

and

d v = e^{x} d x

J_{n} = [e^{x} sin (n x)]_{0}^{π} - n I_{n} = - n I_{n},

since

sin (nπ) = sin 0 = 0

for every integer

n

Put

D_{n} = (- 1)^{n} e^{π} - 1

. The two linear equations from parts (i) and (ii) give

I_{n} = D_{n} + n J_{n}

and

J_{n} = - n I_{n}

, so

(1 + n^{2}) I_{n} = D_{n}

. Therefore

I_{n} = \frac{( - 1 ) ^{n} e ^{π} - 1}{1 + n ^{2}}, J_{n} = \frac{n [ 1 - ( - 1 ) ^{n} e ^{π} ]}{1 + n ^{2}} .

These formulas include

n = 0

, when

I_{0} = e^{π} - 1

and

J_{0} = 0

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