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Laplace transforms of functions tsin²3t

Short answer

For t sin²3t, use sin²3t = (1 - cos6t)/2. L[t]=1/s², L[t cos6t] = (s²-36)/(s²+36)². Result: ½[1/s² - (s²-36)/(s²+36)²] = 54(s²+12)/[ s²(s²+36)²]. For sin2t/t, L[sin2t]=2/(s²+4); integral gives tan⁻¹(2/s) = cot⁻¹(s/2).

Find the Laplace transform of: i) t sin² 3t ii) sin 2t / t.

i) L[t sin² 3t]

sin² 3t = (1 − cos 6t)/2. So t sin² 3t = (t − t cos 6t)/2.

L[t]       = 1/s²
L[t cos 6t] = −d/ds [s/(s² + 36)] = (s² − 36)/(s² + 36)²

L[t sin² 3t] = ½ [ 1/s² − (s² − 36)/(s² + 36)² ]

Simplified: = 54(s² + 12) / [s²(s² + 36)²].

Answer: ½ [1/s² − (s² − 36)/(s² + 36)²]

ii) L[sin 2t / t]

L[sin 2t] = 2/(s² + 4). So

L[sin 2t / t] = ∫ₛ^∞ 2 du/(u² + 4) = [tan⁻¹(u/2)]ₛ^∞ = π/2 − tan⁻¹(s/2)

Answer: tan⁻¹(2/s) = cot⁻¹(s/2)

✓ Answer checked by Claude Opus 5.5 (high effort)

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