Laplace Transform of sin^3t | Laplace of sin cube t

The Laplace transform of sin^3t is equal to 6/[(s2+1)(s2+9)]. In this article, let us learn to find the Laplace of sin cube t. The Laplace transform formula of sin3t is given below:

L{sin3t} = $\dfrac{6}{(s^2+1)(s^2+9)}$.

Find the Laplace Transform of sin3t

Answer: The Laplace transform of sin cube t is L{sin3t} = 6/[(s2+1)(s2+9)].

Proof:

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

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

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

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

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

As L{sin at} = a/(s2+a2), it follows that

L{sin3t} = $\dfrac{3}{4} \dfrac{1}{s^2+1^2}$ – $\dfrac{1}{4} \dfrac{3}{s^2+3^2}$

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

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

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

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

So the Laplace transform of sin^3t is equal to 6/[(s2+1)(s2+9)].

Laplace Transform: Definition, Table, Formulas, Properties

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Laplace transform of cos2t

Laplace transform of sint/t

Laplace transform of t sint

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/[(s2+1)(s2+9)]. It is obtained by applying the Laplace transform on the formula of sin cube t.

Q2: Find L{sin3t}.