---
title: "Laplace transforms of functions te^(−t)sint"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:laplace-transforms-of-functions-te-t-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/laplace-transforms-of-functions-te-t-two
site: BCT Engineering
---

# Laplace transforms of functions te^(−t)sint

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2075 Chaitra, 2+3 marks
- Video: [Laplace's transformation of L [ t² e^t sin4t]. Engineering mathematics by shailesh sj Upadhyay sir](https://www.youtube.com/watch?v=zpN5EXxsiYU) — Rohini educational center of mathematics
- 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-te-t-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Find the Laplace transform of: i) t e^(−t) sin t  ii) (cos 2t − cos 3t)/t.

## Short answer

i) L{e^(−t) sin t} = 1/(( s+1)²+1); multiply by t: −d/ds yields 2(s+1)/(s²+2s+2)². ii) Use L{f/t} = ∫ₛ^∞ F(u) du with F(s)=s/(s²+4)−s/(s²+9) to get ½ ln((s²+9)/(s²+4)).

## Answer

#### i) L{t e^(−t) sin t}
```text
L{sin t} = 1/(s² + 1)
Shift s → s + 1:  L{e^(−t) sin t} = 1/((s + 1)² + 1)
Multiply by t:  −d/ds [((s + 1)² + 1)⁻¹] = 2(s + 1)/((s + 1)² + 1)²
```

So L{t e^(−t) sin t} = 2(s + 1)/(s² + 2s + 2)².

#### ii) L{(cos 2t − cos 3t)/t}
Use L{f(t)/t} = ∫ₛ^∞ F(u) du with F(s) = s/(s² + 4) − s/(s² + 9).

```text
∫ₛ^∞ [u/(u² + 4) − u/(u² + 9)] du = ½ [ln((u² + 4)/(u² + 9))]ₛ^∞
 = −½ ln((s² + 4)/(s² + 9))
 = ½ ln((s² + 9)/(s² + 4))
```

**Answers:** i) 2(s + 1)/(s² + 2s + 2)²; ii) ½ ln((s² + 9)/(s² + 4)).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is L{e^(-t) sin t}?
   A. 1/((s-1)^2 + 1)
   B. 1/(s^2 + 1)
   C. (s+1)/((s+1)^2 + 1)
   D. 1/((s+1)^2 + 1)
   Answer: D. s -> s+1. (Key point: Shifted transform)

2. What is L{t e^(-t) sin t}?
   A. 2(s+1)/(s^2+2s+2)^2
   B. 2s/(s^2+2s+2)^2
   C. (s+1)/(s^2+2s+2)^2
   D. -2(s+1)/(s^2+2s+2)^2
   Answer: A. -d/ds of ((s+1)^2+1)^-1. (Key point: Result for (i))

3. Why combine the two logs before taking the limit u -> infinity?
   A. Each log tends to 1
   B. Each log diverges alone, but their difference tends to 0
   C. The logs are equal
   D. It avoids differentiating
   Answer: B. ln(u^2+4) and ln(u^2+9) both diverge; ln of the ratio tends to ln 1 = 0. (Key point: Combining logs)

## Close variants set in other exams

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