---
title: "Laplace transforms of functions f(t)=(1−cost)/t"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:laplace-transforms-of-functions-f-t"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/laplace-transforms-of-functions-f-t
site: BCT Engineering
---

# Laplace transforms of functions f(t)=(1−cost)/t

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2070 Asar, 5 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-f-t
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Find the Laplace transform of (i) f(t) = (1 − cos t)/t  (ii) f(t) = t e^(−t) sin t.

## Short answer

For (1-cos t)/t, L = ∫_s^∞ (1/u - u/(u^2+1)) du = ½ ln((s^2+1)/s^2). For t e^(-t) sin t, L[t sin t]=2s/(s^2+1)^2, then replace s by s+1 giving 2(s+1)/(( s+1)^2+1)^2.

## Answer

#### (i) L[(1 − cos t)/t]
```text
∫ₛ^∞ [1/u − u/(u² + 1)] du = [ ln( u/√(u² + 1) ) ]ₛ^∞ = −ln( s/√(s² + 1) )
```

**Answer: ½ ln((s² + 1)/s²)**

#### (ii) L[t e^(−t) sin t]
L[sin t] = 1/(s² + 1); L[t sin t] = −d/ds [1/(s²+1)] = 2s/(s²+1)².
Shift s → s + 1:

**Answer: 2(s + 1)/((s + 1)² + 1)²**

## Check yourself

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

1. L[(1 - cos t)/t] uses
   A. -F'(s)
   B. F(s)/s
   C. integral of F(u) from s to infinity
   D. F(s+1)
   Answer: C. Division by t. (Key point: Method (i))

2. L[(1 - cos t)/t] =
   A. (1/2) ln(s^2/(s^2+1))
   B. ln((s^2+1)/s^2)
   C. (1/2) ln((s+1)/s)
   D. (1/2) ln((s^2+1)/s^2)
   Answer: D. -ln(s/sqrt(s^2+1)). (Key point: Result (i))

3. L[t sin t] =
   A. 2s/(s^2+1)^2
   B. 1/(s^2+1)^2
   C. -2s/(s^2+1)^2
   D. 2/(s^2+1)^2
   Answer: A. -d/ds of 1/(s^2+1). (Key point: Step (ii))

4. L[t e^(-t) sin t] =
   A. 2(s-1)/((s-1)^2+1)^2
   B. 2(s+1)/((s+1)^2+1)^2
   C. 2s/((s+1)^2+1)^2
   D. 2(s+1)/((s+1)^2+1)
   Answer: B. Replace s by s + 1. (Key point: Result (ii))

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