---
title: "Laplace transforms of functions sin2tcosh4t"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:laplace-transforms-of-functions-sin2tcosh4t"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/laplace-transforms-of-functions-sin2tcosh4t
site: BCT Engineering
---

# Laplace transforms of functions sin2tcosh4t

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2071 Shrawan, 5 marks
- Video: [Find L[t sin 3t cos 2t] | Maths Naresh E-Class](https://www.youtube.com/watch?v=vq8UQBz3xfU) — maths naresh eclass
- 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-sin2tcosh4t
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Find the Laplace transform of (i) sin 2t cosh 4t  (ii) t e^(2t) sin 4t.

## Short answer

Expres cosh4t as (e^(4t)+e^(-4t))/2, apply shifting to L[sin2t]=2/(s^2+4) giving 1/(( s-4)^2+4) +1/(( s+4)^2+4), simplify to 2(s^2+20)/(s^4-24s^2+400). For t e^(2t) sin4t, L[t sin4t]=8s/(s^2+16)^2, shift s→s-2 to get 8(s-2)/(( s-2)^2+16)^2.

## Answer

#### (i) L[sin 2t cosh 4t]
cosh 4t = (e^(4t) + e^(−4t))/2 and L[sin 2t] = 2/(s² + 4). By the first shifting theorem,

```text
L[sin 2t cosh 4t] = ½ [ 2/((s − 4)² + 4) + 2/((s + 4)² + 4) ]
                  = 1/((s − 4)² + 4) + 1/((s + 4)² + 4)
```

Combine: the denominators multiply to (s² + 20)² − 64s² = s⁴ − 24s² + 400, and the numerator is 2(s² + 20).

**Answer: 2(s² + 20)/(s⁴ − 24s² + 400)**

#### (ii) L[t e^(2t) sin 4t]
L[sin 4t] = 4/(s² + 16).
L[t sin 4t] = −d/ds [4/(s² + 16)] = 8s/(s² + 16)².
Shift s → s − 2:

**Answer: 8(s − 2)/((s − 2)² + 16)²**

## Check yourself

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

1. For sin 2t cosh 4t we use
   A. the integral theorem
   B. convolution
   C. differentiation of F
   D. cosh 4t = (e^(4t)+e^(-4t))/2 and the first shift
   Answer: D. Shifts s by -+4. (Key point: Approach (i))

2. L[sin 2t cosh 4t] =
   A. 2(s^2+20)/(s^4 - 24s^2 + 400)
   B. 2(s^2-20)/(s^4 - 24s^2 + 400)
   C. 2(s^2+20)/(s^4 + 24s^2 + 400)
   D. (s^2+20)/(s^4 - 24s^2 + 400)
   Answer: A. Denominator (s^2+20)^2 - 64s^2. (Key point: Result (i))

3. L[t sin 4t] =
   A. 4/(s^2+16)^2
   B. 8s/(s^2+16)^2
   C. -8s/(s^2+16)^2
   D. 8/(s^2+16)^2
   Answer: B. -d/ds of 4/(s^2+16). (Key point: Part (ii) step)

4. L[t e^(2t) sin 4t] =
   A. 8(s+2)/((s+2)^2+16)^2
   B. 8(s-2)/((s-2)^2+16)
   C. 8(s-2)/((s-2)^2+16)^2
   D. 8s/((s-2)^2+16)^2
   Answer: C. Replace s by s - 2. (Key point: Result (ii))

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