The Laplace transform of sin t is equal to 1/(s^{2}+1), and it is denoted by L{sint}. Here we will find the Laplace of sint by the definition of Laplace transforms.

The formula of the Laplace of sint is given by

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

## Laplace transform of sint

By definition, the Laplace transform of f(t) is given by

L{f(t)} = $\int_0^\infty$ e^{-st} f(t) dt

Let f(t) = sint.

Then the Laplace of sin t is equal to

L{sin t} = $\int_0^\infty$ e^{-st} sin t dt

= $\Big[\dfrac{e^{-st}}{s^2+1}(-s \sin t – \cos t) \Big]_0^\infty$, using the integration formulas listed here.

= lim_{b→∞} $\Big[\dfrac{e^{-sb}}{s^2+1}(-s \sin b – \cos b) \Big]$ $+ \dfrac {e^{0} (s \sin 0 + \cos 0) } {s^2 + 1}$

= $0+ \dfrac{1}{s^2 + 1}$ as we know sin0=0 and cos 0=1.

= $\dfrac{1} {s^2 +1}$

So the Laplace transform of sint is equal to 1/(s^{2}+1) and this is proved by the definition of Laplace transforms. That is,

L{sint} = 1/(s^{2}+1). |

Related Topics:

Laplace transform of unit step function

Laplace transform of Dirac delta function

## FAQs

**Q1: What is the Laplace transform of sint?**

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

**Q2: Find L{sin t}.**

Answer: L{sin t} = 1/(s^{2}+1).