The function t cos(t) is the product of t and cosine of t. The Laplace transform of tcos(t) is (s^{2}-1)/(s^{2}+1)^{2}. In this article, we will find the Laplace transform of both tcos(t) and tcos(at).

Table of Contents

## What is the Laplace Transform of t cos(t)?

**Answer:** The Laplace transform of t cos t is (s^{2}-1)/(s^{2}+1)^{2}.

*Proof:*

We know that the Laplace transform of a function f(t) multiplied by t, denoted by L{t f(t)}, is given by the following multiplication by t Laplace transform formula:

$L\{t f(t)\} = – \dfrac{d}{ds}(F(s))$, where L{f(t)}=F(s) **…(∗)**

**Step 1:** Put f(t) = cos(t) in the above formula.

∴ F(s) = L{f(t)} = L{cos(t)} = s/(s^{2}+1)

**Step 2:** Now, by the formula **(∗)**, the Laplace transform of tcos(t) is equal to

$L\{t\cos(t)\} = – \dfrac{d}{ds}\left(\dfrac{s}{s^2+1}\right)$

**Step 3:** Applying the quotient rule of derivatives, we obtain that

$L\{t\cos(t)\}$ $= – \dfrac{(s^2+1)\frac{d}{ds}(s)-s \frac{d}{ds}(s^2+1)}{(s^2+1)^2}$

$= – \dfrac{(s^2+1)\cdot 1-s \cdot 2s}{(s^2+1)^2}$

$= – \dfrac{s^2+1-2s^2}{(s^2+1)^2}$

$= \dfrac{s^2-1}{(s^2+1)^2}$.

So the Laplace transform of tcos t is (s^{2}-1)/(s^{2}+1)^{2}.

Find the Laplace transform of t cos(t).Summary:L{t cos t} = (s ^{2}-1)/(s^{2}+1)^{2}. |

**Also Read:**

Laplace transform of t: | 1/s^{2} |

Laplace transform of sin t: | 1/(s^{2}+1) |

Laplace transform of sin(t)/t: | tan^{-1}(1/s) |

Laplace transform of cos t: | s/(s^{2}+1) |

Laplace transform of cos(t)/t: | Does Not Exist |

Laplace transform of e^{-t}: | 1/(s+1) |

Laplace transform of 1: | 1/s |

## What is the Laplace Transform of t cos(at)?

**Answer:** The Laplace transform of t cos at is (s^{2}-a^{2})/(s^{2}+a^{2})^{2}.

*Proof:*

In the above formula **(∗)**, we put f(t) = t cos(at). As L{cos at} = s/(s^{2}+a^{2}), the Laplace transform of t cos(at) by the above formula **(∗)** will be equal to

$L\{t\cos(at)\} = – \dfrac{d}{ds}\left(\dfrac{s}{s^2+a^2}\right)$

$= – \dfrac{(s^2+a^2)\frac{d}{ds}(s)-s \frac{d}{ds}(s^2+a^2)}{(s^2+a^2)^2}$

$= – \dfrac{(s^2+a^2)\cdot 1-s \cdot 2s}{(s^2+a^2)^2}$

$= – \dfrac{s^2+a^2-2s^2}{(s^2+a^2)^2}$

$= \dfrac{s^2-a^2}{(s^2+a^2)^2}$.

So the Laplace transform of tcos at is (s^{2}-a^{2})/(s^{2}+a^{2})^{2}.

## FAQs

**Q1: t cos(t) Laplace transform.**

Answer: The Laplace transform of the product tcost is (s^{2}-1)/(s^{2}+1)^{2}, that is, L{t cos t} = (s^{2}-1)/(s^{2}+1)^{2}.

**Q2: Find the Laplace transform formula of t cos(at).**

Answer: The Laplace transform of the product tcosat is (s^{2}-a^{2})/(s^{2}+a^{2})^{2}, that is, L{t cos at} = (s^{2}-a^{2})/(s^{2}+a^{2})^{2}.

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.