Laplace Transform of e^-t sint | Find L{e^-t sint}

The Laplace transform of e^-t sint, that is, L{e-t sint} is equal to 1/[(s+1)2+1]. Note e-t sint is a product of two functions, and its Laplace is calculated using the first shifting property of Laplace transforms.

The Laplace formula of e-t sint is given below.

L{e-t sint} = 1/[(s+1)2+1].
Laplace Transform of e^-t sint

Table of Contents

Laplace of e-t sint

Answer: The Laplace of e-t sint is 1/[(s+1)2+1].

Step 1: First, we calculate the Laplace of sint using the formula of L{sinat}.

L{sint} = $\dfrac{1}{s^2+1}$.

Step 2: Now use the first shifting property of Laplace transforms which says that if L{f(t)} = F(s), then

L{eat f(t)} = F(s-a).

Step 3: In our case, a=-1 and f(t) = sint. So F(s) = $\dfrac{1}{s^2+1}$.

Thus, L{e-t sint} = F(s+1) = $\dfrac{1}{(s+1)^2+1}$.

That is, L{e-t sint} = $\dfrac{1}{s^2+2s+2}$.

So the Laplace transform of e-t sint is equal to 1/(s2+2s+2), and this is obtained by using the first shifting property of Laplace transforms.

More Laplace Transforms:

Laplace transform of t cost

Laplace transform of t sint

Laplace transform of tet

Laplace transform of (1-sint)/t

Laplace transform of (1-cost)/t

Laplace transform of (1-et)/t

FAQs

Q1: What is the Laplace transform of e-t sint?

Answer: The Laplace transform of e-t sint is equal to 1/(s2+2s+2).

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