 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/ii/solution

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

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