 Solution

ID: 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/i/solution

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}

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