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]. |

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## Laplace of e^{-t} sint

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

**Explanation:**

**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{e^{at }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/(s^{2}+2s+2), and this is obtained by using the first shifting property of Laplace transforms.

More Laplace Transforms:

Laplace transform of (1-sint)/t

Laplace transform of (1-cost)/t

Laplace transform of (1-e^{t})/t

## FAQs

**Q1: What is the Laplace transform of e**

^{-t}sint?Answer: The Laplace transform of e^{-t} sint is equal to 1/(s^{2}+2s+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.