# Laplace Transform of sin^2t | Laplace of sin square t

The Laplace transform of sin^2t is equal to 1/2s – s/(2s2+8). In this article, we will learn how to find the Laplace of sin square t. The formula of the Laplace of sin2t is given below:

L{sin2t} = 1/2s – s/(2s2+8).

## Find the Laplace Transform of sin2t

Answer: The Laplace transform of sin square t is 1/2s – s/(2s2+8).

Proof:

It is known from trigonometry that 1-cos2t=2sin2t. Therefore, we have

sin2t = $\frac{1}{2}$(1-cos2t)

Now, we take the Laplace transforms on both sides of the above equation. By doing so, we get that

L{sin2t} = L{$\frac{1}{2}$(1-cos2t)}.

Applying the linearity property of Laplace transforms,

L{sin2t} = $\frac{1}{2}$ (L{1} – L{cos2t}).

We know that the Laplace Transform of 1 is 1/s and the Laplace transform of cos 2t is s/(s2+4). Hence, it follows that

L{sin2t} = $\dfrac{1}{2} \left(\dfrac{1}{s}-\dfrac{s}{s^2+4} \right)$

Simplifying we obtain that

L{sin2t} = $\dfrac{1}{2s}-\dfrac{s}{2(s^2+4)}$

Therefore, the Laplace transform of sin^2t is equal to 1/2s – s/(2s2+8).

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## FAQs

Q1: What is the Laplace transform of sin^2t?

Answer: The Laplace transform of sin^2t is equal to 1/2s – s/(2s2+8). That is, L{sin2t} = 1/2s – s/(2s2+8).

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