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

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

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2078 Kartik, 1+1.5+2.5 marks
- Video: [L[te^-t cosht] | Laplace Transform | L1k,280](https://www.youtube.com/watch?v=54IdbcUXmyw) — Maths Pulse - Chinnaiah Kalpana
- 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
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Define the Laplace transform of f(t). Find the Laplace transform of: a) t e^(−t) cosh t  b) sin t sin 5t / t.

## Short answer

L{f(t)} = ∫0∞ e^{-st}f(t)dt. a) t e^{-t} cosh t = (t + t e^{-2t})/2, transform is (1/2)[1/s^2 + 1/(s+2)^2]. b) L{sin t sin 5t / t} = (1/4) ln((s^2+36)/(s^2+16)).

## Answer

#### Definition
L{f(t)} = F(s) = ∫₀^∞ e^(−st) f(t) dt, for those s for which the integral converges.

#### a) L{t e^(−t) cosh t}
Since e^(−t) cosh t = (1 + e^(−2t))/2, we have t e^(−t) cosh t = (t + t e^(−2t))/2.

```text
L{t} = 1/s²,   L{t e^(−2t)} = 1/(s + 2)²
L{t e^(−t) cosh t} = ½ [1/s² + 1/(s + 2)²]        (s > 0)
```

#### b) L{sin t sin 5t / t}
Use sin t sin 5t = ½(cos 4t − cos 6t). Then L{f(t)/t} = ∫ₛ^∞ F(u) du with F(s) = ½[s/(s² + 16) − s/(s² + 36)].

```text
∫ₛ^∞ ½ [u/(u² + 16) − u/(u² + 36)] du = ¼ [ln((u² + 16)/(u² + 36))]ₛ^∞
 = 0 − ¼ ln((s² + 16)/(s² + 36))
 = ¼ ln((s² + 36)/(s² + 16))
```

**Answers:** a) ½[1/s² + 1/(s + 2)²]; b) ¼ ln((s² + 36)/(s² + 16)).

## 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 -infinity to infinity of e^(-st) f(t) dt
   B. integral from 0 to infinity of e^(st) f(t) dt
   C. integral from 0 to 1 of e^(-st) f(t) dt
   D. integral from 0 to infinity of e^(-st) f(t) dt
   Answer: D. 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. What is L{t e^(-t) cosh t}?
   A. (1/2)[1/s^2 + 1/(s+2)^2]
   B. (1/2)[1/s^2 + 1/(s-2)^2]
   C. (1/2)[1/s + 1/(s+2)]
   D. 1/(s+1)^2
   Answer: A. e^(-t) cosh t = (1 + e^(-2t))/2, so t times it gives (t + t e^(-2t))/2. (Key point: Part (a))

3. sin t sin 5t equals what?
   A. (1/2)(cos 4t + cos 6t)
   B. (1/2)(cos 4t - cos 6t)
   C. (1/2)(sin 4t - sin 6t)
   D. (1/2)(cos 6t - cos 4t)
   Answer: B. sin A sin B = (1/2)[cos(A-B) - cos(A+B)]. (Key point: Product to sum)

4. What is L{sin t sin 5t / t}?
   A. (1/2) ln((s^2+36)/(s^2+16))
   B. (1/4) ln((s^2+16)/(s^2+36))
   C. (1/4) ln((s^2+36)/(s^2+16))
   D. (1/4) ln((s^2+36)(s^2+16))
   Answer: C. Integrating (1/2)[u/(u^2+16) - u/(u^2+36)] gives (1/4) ln ratio, then 0 minus it flips the ratio. (Key point: Part (b))

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