The Laplace transform of sin^3t is equal to 6/[(s^{2}+1)(s^{2}+9)]. In this article, let us learn to find the Laplace of sin cube t. The Laplace transform formula of sin^{3}t is given below:

L{sin^{3}t} = $\dfrac{6}{(s^2+1)(s^2+9)}$.

Table of Contents

## Find the Laplace Transform of sin^{3}t

**Answer:** The Laplace transform of sin cube t is L{sin^{3}t} = 6/[(s^{2}+1)(s^{2}+9)].

*Proof:*

From the theory of trigonometry, we know that sin3t=3sint – 4sin^{3}t. Hence, sin^{3}t can be written as follows

sin^{3}t = $\dfrac{1}{4}$(3sint – sin3t).

⇒ sin^{3}t = $\dfrac{3}{4}$ sint – $\dfrac{1}{4}$ sin3t

Taking Laplace transforms on both sides and using the linearity property of Laplace transforms, we obtain that

L{sin^{3}t} = $\dfrac{3}{4}$ L{sint} – $\dfrac{1}{4}$ L{sin3t}.

As L{sin at} = a/(s^{2}+a^{2}), it follows that

L{sin^{3}t} = $\dfrac{3}{4} \dfrac{1}{s^2+1^2}$ – $\dfrac{1}{4} \dfrac{3}{s^2+3^2}$

= $\dfrac{3}{4} \Big[\dfrac{1}{s^2+1} – \dfrac{1}{s^2+9} \Big]$

= $\dfrac{3}{4} \dfrac{s^2+9-s^2-1}{(s^2+1)(s^2+9)}$

= $\dfrac{3}{4} \dfrac{8}{(s^2+1)(s^2+9)}$

= $\dfrac{6}{(s^2+1)(s^2+9)}$

**Final Answer:**

So the Laplace transform of sin^3t is equal to 6/[(s^{2}+1)(s^{2}+9)].

Have You Read These?

Laplace Transform: Definition, Table, Formulas, Properties

Laplace transform of sin2t sin3t

Laplace transform of sint sin2t sin3t

## FAQs

**Q1: What is the Laplace transform of sin^3t?**

Answer: The Laplace transform of sin^3t is equal to 6/[(s^{2}+1)(s^{2}+9)]. It is obtained by applying the Laplace transform on the formula of sin cube t.

**Q2: Find L{sin**

^{3}t}.Answer: L{sin^{3}t} = 6/[(s^{2}+1)(s^{2}+9)].

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.