We want to find the Laplace transform of $F(t) = \sin(kt)$.

1. Problem Description

We want to find the Laplace transform of $F(t) = \sin(kt)$.

2. Solution Steps

The Laplace transform of a function $F(t)$ is defined as:

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

So, we need to evaluate the integral:

$L\{\sin(kt)\} = \int_0^\infty e^{-st} \sin(kt) dt$

We can solve this integral using integration by parts twice. Let $u = \sin(kt)$ and $dv = e^{-st} dt$. Then $du = k \cos(kt) dt$ and $v = -\frac{1}{s}e^{-st}$. So,

$\int_0^\infty e^{-st} \sin(kt) dt = \left[-\frac{1}{s}e^{-st} \sin(kt)\right]_0^\infty + \frac{k}{s} \int_0^\infty e^{-st} \cos(kt) dt$

$= 0 + \frac{k}{s} \int_0^\infty e^{-st} \cos(kt) dt$

Now we integrate by parts again. Let $u = \cos(kt)$ and $dv = e^{-st} dt$. Then $du = -k \sin(kt) dt$ and $v = -\frac{1}{s} e^{-st}$.

$\int_0^\infty e^{-st} \cos(kt) dt = \left[-\frac{1}{s} e^{-st} \cos(kt)\right]_0^\infty - \frac{k}{s} \int_0^\infty e^{-st} \sin(kt) dt$

$= \frac{1}{s} - \frac{k}{s} \int_0^\infty e^{-st} \sin(kt) dt$

Substitute this back into the original equation:

$\int_0^\infty e^{-st} \sin(kt) dt = \frac{k}{s} \left[\frac{1}{s} - \frac{k}{s} \int_0^\infty e^{-st} \sin(kt) dt\right]$

$\int_0^\infty e^{-st} \sin(kt) dt = \frac{k}{s^2} - \frac{k^2}{s^2} \int_0^\infty e^{-st} \sin(kt) dt$

Let $I = \int_0^\infty e^{-st} \sin(kt) dt$. Then

$I = \frac{k}{s^2} - \frac{k^2}{s^2} I$

$I + \frac{k^2}{s^2} I = \frac{k}{s^2}$

$I \left(1 + \frac{k^2}{s^2}\right) = \frac{k}{s^2}$

$I \left(\frac{s^2 + k^2}{s^2}\right) = \frac{k}{s^2}$

$I = \frac{k}{s^2 + k^2}$

Therefore, $L\{\sin(kt)\} = \frac{k}{s^2 + k^2}$.

3. Final Answer

$L\{\sin(kt)\} = \frac{k}{s^2 + k^2}$