---
title: "Laplace transforms of functions te^(−4t)sin3t"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:laplace-transforms-of-functions-te-4t"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/laplace-transforms-of-functions-te-4t
site: BCT Engineering
---

# Laplace transforms of functions te^(−4t)sin3t

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2079 Bhadra, 1+2+2 marks
- Video: [Laplace Transform of te^-2tsin2t || Laplace Transform example](https://www.youtube.com/watch?v=YcfphxmyObY) — Let's Unlock Maths
- 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-4t
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Define the Laplace transform of a function f(t). Find the Laplace transform of a) t e^(−4t) sin 3t  b) (1 − eᵗ)/t.

## Short answer

L{f(t)} = ∫0∞ e^{-st} f(t) dt for s where integral converges. a) L{t e^{-4t} sin 3t} = 6(s+4)/(( (s+4)^2 + 9 )^2. b) L{(1 - e^t)/t} = ln((s-1)/s).

## Answer

#### Definition
For a function f(t) defined for t ≥ 0, the Laplace transform is

L{f(t)} = F(s) = ∫₀^∞ e^(−st) f(t) dt,

provided the integral converges (for s large enough).

#### a) L{t e^(−4t) sin 3t}
```text
L{sin 3t} = 3/(s² + 9)
First shifting (s → s + 4): L{e^(−4t) sin 3t} = 3/((s + 4)² + 9)
Multiply by t: L{t f(t)} = −dF/ds
 = −d/ds [3((s + 4)² + 9)⁻¹]
 = 6(s + 4)/((s + 4)² + 9)²
```

#### b) L{(1 − eᵗ)/t}
Use L{f(t)/t} = ∫ₛ^∞ F(u) du with F(s) = 1/s − 1/(s − 1).

```text
∫ₛ^∞ [1/u − 1/(u − 1)] du = [ln(u/(u − 1))]ₛ^∞ = 0 − ln(s/(s − 1)) = ln((s − 1)/s)     (s > 1)
```

**Answers:** a) 6(s + 4)/((s + 4)² + 9)²; b) ln((s − 1)/s).

## Check yourself

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

1. Which integral defines L{f(t)}?
   A. integral from 0 to infinity of e^(-st) f(t) dt
   B. integral from -infinity to infinity of e^(-st) f(t) dt
   C. integral from 0 to infinity of e^(st) f(t) dt
   D. integral from 0 to 1 of e^(-st) f(t) dt
   Answer: A. L{f(t)} = F(s) is the integral over t from 0 to infinity of e^(-st) f(t), provided it converges for large enough s. (Key point: Definition of Laplace transform)

2. For t e^(-4t) sin 3t, what is the first-shift result?
   A. 3/((s-4)^2 + 9)
   B. L{e^(-4t) sin 3t} = 3/((s+4)^2 + 9)
   C. 3(s+4)/((s+4)^2 + 9)
   D. 3/(s^2 + 9)
   Answer: B. e^(-4t) shifts s to s+4; then multiply by t via -d/ds. (Key point: Order of operations in part (a))

3. What is L{t e^(-4t) sin 3t}?
   A. 6(s-4)/((s-4)^2 + 9)^2
   B. 3/((s+4)^2 + 9)^2
   C. 6(s+4)/((s+4)^2 + 9)^2
   D. 6s/(s^2 + 9)^2
   Answer: C. -d/ds [3((s+4)^2+9)^-1] = 6(s+4)/((s+4)^2+9)^2. (Key point: Result for part (a))

4. What is L{(1 - e^t)/t}?
   A. ln(s/(s-1))
   B. ln(s(s-1))
   C. ln(s-1) + ln s
   D. ln((s-1)/s)
   Answer: D. The integral equals 0 - ln(s/(s-1)) = ln((s-1)/s), valid for s > 1. (Key point: Result for part (b))

## Close variants set in other exams

- [Laplace transforms of functions te^(3t)sin3t](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Alaplace-transforms-of-functions-te-3t) — Mathematics III, chapter 3, asked 9×
- [Laplace transforms of functions tsin²3t](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Alaplace-transforms-of-functions-tsin23t) — Mathematics III, chapter 3, asked 9×
- [Laplace transforms of functions t²sin2t](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Alaplace-transforms-of-functions-t2sin2t-two) — Mathematics III, chapter 3, asked 9×

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