 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-4/12a/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

For one period,

\int_{0}^{2 π} e^{- s t} h (t) d t = \int_{0}^{π} e^{- s t} sin t d t = \frac{1 + e ^{- π s}}{s ^{2} + 1} .

Part (b), followed by cancellation of

1 + e^{- π s}

, gives

L {h} (s) = \frac{1}{( s ^{2} + 1 ) ( 1 - e ^{- π s} )} .

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