--- title: "Laplace transforms of functions tsin²3t" subject: "Mathematics III" chapter: 3 key: "semester_3/mathematics_iii:laplace-transforms-of-functions-tsin23t" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/laplace-transforms-of-functions-tsin23t site: BCT Engineering --- # Laplace transforms of functions tsin²3t Mathematics III · Chapter 3: Laplace Transform - Asked in: 2072 Chaitra, 2+3 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/laplace-transforms-of-functions-tsin23t - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3 ## Question Find the Laplace transform of: i) t sin² 3t ii) sin 2t / t. ## 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). ## Answer #### i) L[t sin² 3t] sin² 3t = (1 − cos 6t)/2. So t sin² 3t = (t − t cos 6t)/2. ```text L[t] = 1/s² L[t cos 6t] = −d/ds [s/(s² + 36)] = (s² − 36)/(s² + 36)² ``` ```text 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 ```text 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)** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. sin^2 3t = ? A. (1 + cos 6t)/2 B. (1 - cos 6t)/2 C. (1 - cos 3t)/2 D. (1 - sin 6t)/2 Answer: B. Half-angle. (Key point: Identity) 2. What is L{t cos 6t}? A. (36-s^2)/(s^2+36)^2 B. 2s/(s^2+36)^2 C. (s^2-36)/(s^2+36)^2 D. s/(s^2+36) Answer: C. -d/ds s/(s^2+36). (Key point: L[t cos 6t]) 3. What is L{t sin^2 3t}? A. (1/2)[1/s^2 + (s^2-36)/(s^2+36)^2] B. 1/s^2 - (s^2-36)/(s^2+36)^2 C. (1/2)[1/s - s/(s^2+36)] D. (1/2)[1/s^2 - (s^2-36)/(s^2+36)^2] Answer: D. Combine (t - t cos 6t)/2. (Key point: Result for (i)) 4. What is L{sin 2t / t}? A. tan^-1(2/s) B. tan^-1(s/2) C. 2/(s^2+4) D. pi/2 + tan^-1(s/2) Answer: A. pi/2 - tan^-1(s/2). 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