# Laplace Transform of sin(at) | Laplace Transform of sin ωt

The Laplace transform of sine function sin(at) is a/(s2+a2) and the Laplace transform of sin(t) is 1/(s2+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/(s2+a2).

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

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

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

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

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

Thus, the Laplace transform of sin(at) is a/(s2+a2) and this is obtained by the definition of Laplace transforms.

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

L{eiat} = $\dfrac{1}{s-ia}$

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

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

⇒ 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/(s2+a2).

Laplace transform of cos(at): The Laplace transform of cosat is s/(s2+a2).

Laplace transform of tn: The Laplace transform of tn is n!/sn+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/(s2+a2). Thus, putting a=1, we will get the Laplace transform of sin(t) which is

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

L{$f^{\prime\prime}(t)$} = s2 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} = s2L{sin t}-s⋅0-1

⇒ – L{sin t} = s2 L{sin t}-1

⇒ 1 = (s2+1) L{sin t}

⇒ L{sin t} = 1/(s2+1).

So the Laplace transform of sint is 1/(s2+1) which is obtained by the Laplace transform of derivatives.

## FAQs

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

Answer: The Laplace transform formula of sin t is 1/(s2+1).

Q2: sin(at) Laplace transform?

Answer: The Laplace transform of sin at is a/(s2+a2).

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