The Laplace transform of sine function sin(at) is a/(s^{2}+a^{2}) and the Laplace transform of sin(t) is 1/(s^{2}+1). Here, we will learn how to find out the Laplace transform of sine functions.

First, recall the Laplace transform definition: The Laplace transform of f(t), denoted by L{f(t)} or F(s), is defined as follows:

F(s) = L{f(t)} = $\int_0^\infty$ e^{-st} f(t) dt **…(I)**

## Laplace Transform of Sine Function

**Theorem:** The Laplace transform of sin at is

L{sin at} = a/(s^{2}+a^{2}).

*Proof:*

We will find the Laplace transform of sin at by the definition. In the above definition **(I)**, we put f(t) = sin at.

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

Using an application of integration by parts formula, we have

L{sin at} = $[\dfrac{e^{-st}}{s^2+a^2}(-s \sin at -a \cos at)]_0^\infty$

= lim_{T→∞} $[\dfrac{e^{-st}}{s^2+a^2}(-s \sin at -a \cos at)]_0^T$

= lim_{T→∞} $[\dfrac {s \sin 0 + a \cos 0} {s^2 + a^2}$ $- \dfrac {e^{-s T} (s \sin a T + a \cos a T) } {s^2 + a^2}]$

= $\dfrac {s \cdot 0 + a \cdot 1} {s^2 + a^2}-0$ as sin0=0 and cos 0=1.

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

Thus, the Laplace transform of sin(at) is a/(s^{2}+a^{2}) and this is obtained by the definition of Laplace transforms.

Find the Laplace transform of sin(at).Summary:L{sin at} = a/(s ^{2}+a^{2}) |

**Alternative Proof:** We will now find the Laplace transform of sin at using the Laplace transform formula of exponential functions. Note that

L{e^{iat}} = $\dfrac{1}{s-ia}$

= $\dfrac{s+ia}{(s-ia)(s+ia)}$, multiplying both numerator and denominator by (s-ia).

Thus, L{e^{iat}} = $\dfrac{s+ia}{s^2-i^2a^2}$

⇒ L{cos at + sin at} = $\dfrac{s}{s^2+a^2}+i\dfrac{a}{s^2+a^2}$

⇒ L{cos at} + L{sin at} = $\dfrac{s}{s^2+a^2}+i\dfrac{a}{s^2+a^2}$ by the linearity property of Laplace transform.

Now, comparing the real part and the imaginary part of both sides, we obtain that

L{sin at} = a/(s^{2}+a^{2}).

**Read Also:**

**Laplace transform of cos(at)**: The Laplace transform of cosat is s/(s^{2}+a^{2}).

**Laplace transform of t ^{n}**: The Laplace transform of t

^{n}is n!/s

^{n+1}..

**Inverse Laplace transform of 1**: The inverse Laplace transform of 1 is the Dirac delta function δ(t).

## Laplace Transform of sin t

In the above, we have shown that the Laplace transform of sin at is equal to a/(s^{2}+a^{2}). Thus, putting a=1, we will get the Laplace transform of sin(t) which is

L{sin t} = 1/(s^{2}+1) |

Let us now find the Laplace transform of sin t using the Laplace transform of second derivatives. It says that

L{$f^{\prime\prime}(t)$} = s^{2} L{f(t)}-sf(0)-$f^\prime(0)$.

Put f(t) = sin t in the above formula. We have f$^{\prime}$(t) = cos t, f$^{\prime\prime}$(t)= -sin t, f(0)=sin0 =0, f$^{\prime}$(0)=cos0=1. Thus, we have

L{- sin t} = s^{2}L{sin t}-s⋅0-1

⇒ – L{sin t} = s^{2} L{sin t}-1

⇒ 1 = (s^{2}+1) L{sin t}

⇒ L{sin t} = 1/(s^{2}+1).

So the Laplace transform of sint is 1/(s^{2}+1) which is obtained by the Laplace transform of derivatives.

## Laplace Transform of sinat Formula

Sine Function | Laplace Transform |

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

sin2t | L{sin2t}= 2/(s^{2}+4) |

sin3t | L{sin3t}= 3/(s^{2}+9) |

sin4t | L{sin4t}= 4/(s^{2}+16) |

sin5t | L{sin5t}= 5/(s^{2}+25) |

sin6t | L{sin6t}= 6/(s^{2}+36) |

sin7t | L{sint7}= 7/(s^{2}+49) |

sin8t | L{sin8t}= 8/(s^{2}+64) |

sin9t | L{sin9t}= 9/(s^{2}+81) |

sin10t | L{sin10t}= 10/(s^{2}+100) |

## FAQs

**Q1: What is the Laplace transform formula of sin t?**

Answer: The Laplace transform formula of sin t is 1/(s^{2}+1).

**Q2: sin(at) Laplace transform?**

Answer: The Laplace transform of sin at is a/(s^{2}+a^{2}).