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

# Laplace transforms of functions

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2072 Kartik, 2.5+2.5 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-three
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Find the Laplace transform of the following: a) (cos 2t − cos 3t)/t  b) sin³ 2t.

## Short answer

For (cos2t - cos3t)/t, use ∫_s^∞ (u/(u^2+4) - u/(u^2+9)) du = ½ ln((s^2+9)/(s^2+4)). For sin^3 2t, rewrite using sin3θ identity: sin^3 2t = (3 sin2t - sin6t)/4, L = 48/(( s^2+4)(s^2+36)).

## Answer

#### a) L[(cos 2t − cos 3t)/t]
```text
∫ₛ^∞ [u/(u² + 4) − u/(u² + 9)] du = ½ [ ln((u² + 4)/(u² + 9)) ]ₛ^∞
                                  = −½ ln((s² + 4)/(s² + 9))
```

**Answer: ½ ln((s² + 9)/(s² + 4))**

#### b) L[sin³ 2t]
Use sin 3θ = 3 sin θ − 4 sin³ θ, so sin³ θ = (3 sin θ − sin 3θ)/4. With θ = 2t:

```text
sin³ 2t = (3 sin 2t − sin 6t)/4
L = ¼ [ 3·2/(s² + 4) − 6/(s² + 36) ]
  = (6/4) [ (s² + 36 − s² − 4) / ((s² + 4)(s² + 36)) ]
  = (3/2)·32 / ((s² + 4)(s² + 36))
```

**Answer: 48/((s² + 4)(s² + 36))**

## Check yourself

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

1. (cos 2t - cos 3t)/t is handled via
   A. differentiation of F
   B. F(s)/s
   C. shifting
   D. integral of F(u) from s to infinity
   Answer: D. Division by t. (Key point: Division by t)

2. L[(cos 2t - cos 3t)/t] =
   A. (1/2) ln((s^2+9)/(s^2+4))
   B. (1/2) ln((s^2+4)/(s^2+9))
   C. ln((s^2+9)/(s^2+4))
   D. (1/2) ln((s+3)/(s+2))
   Answer: A. Evaluate the log antiderivative. (Key point: Result (a))

3. sin^3 theta =
   A. (3 sin theta + sin 3 theta)/4
   B. (3 sin theta - sin 3 theta)/4
   C. (sin 3 theta - 3 sin theta)/4
   D. 3 sin theta - 4 sin 3 theta
   Answer: B. From sin 3x = 3 sin x - 4 sin^3 x. (Key point: Identity)

4. L[sin^3 2t] =
   A. 24/((s^2+4)(s^2+36))
   B. 48/(s^2+36)
   C. 48/((s^2+4)(s^2+36))
   D. 12/((s^2+4)(s^2+36))
   Answer: C. (1/4)[6/(s^2+4) - 6/(s^2+36)]. (Key point: Result (b))

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