$\displaystyle \mathcal{L} \left\{f(t)\right\} = \int_0^\infty e^{-st} f(t) \, dt$
$\displaystyle \mathcal{L} (\sin bt) = \int_0^\infty e^{-st} \sin bt \, dt$
For $\displaystyle \int_0^\infty e^{-st} \sin bt \, dt$.
Using integration by parts: $\displaystyle \int u\,dv = uv - \int v\, du$. Let
$u = e^{-st}$ $du = -se^{-st} \, dt$
$dv = \sin bt \, dt$ $v = -\dfrac{1}{b} \cos bt$
$\displaystyle \int_0^\infty e^{-st} \sin bt \, dt = \left[ -\dfrac{1}{b}e^{-st} \cos bt \right]_0^\infty - \dfrac{s}{b} \int_0^\infty e^{-st} \cos bt \, dt$
Using integration by parts again. Let
$u = e^{-st}$ $du = -se^{-st} \, dt$
$dv = \cos bt \, dt$ $v = \dfrac{1}{b} \sin bt$
$\displaystyle \int_0^\infty e^{-st} \sin bt \, dt = \left[ -\dfrac{1}{b}e^{-st} \cos bt \right]_0^\infty - \dfrac{s}{b} \left[ \dfrac{1}{b}e^{-st} \sin bt + \dfrac{s}{b} \int e^{-st} \sin bt \, dt \right]_0^\infty$
$\displaystyle \int_0^\infty e^{-st} \sin bt \, dt = \left[ -\dfrac{1}{b}e^{-st} \cos bt \right]_0^\infty - \left[ \dfrac{s}{b^2} e^{-st} \sin bt \right]_0^\infty - \dfrac{s^2}{b^2} \int_0^\infty e^{-st} \sin bt \, dt$
$\displaystyle \int_0^\infty e^{-st} \sin bt \, dt \, + \, \dfrac{s^2}{b^2} \int_0^\infty e^{-st} \sin bt \, dt = \left[ -\dfrac{1}{b}e^{-st} \cos bt \right]_0^\infty - \left[ \dfrac{s}{b^2} e^{-st} \sin bt \right]_0^\infty$
$\displaystyle b^2 \int_0^\infty e^{-st} \sin bt \, dt \, + \, s^2 \int_0^\infty e^{-st} \sin bt \, dt = \left[ -be^{-st} \cos bt \right]_0^\infty - \left[ se^{-st} \sin bt \right]_0^\infty$
$\displaystyle (b^2 + s^2)\int_0^\infty e^{-st} \sin bt \, dt = \left[ -\dfrac{b \cos bt}{e^{st}} \right]_0^\infty - \left[ \dfrac{s \sin bt}{e^{st}} \right]_0^\infty$
$\displaystyle (s^2 + b^2)\int_0^\infty e^{-st} \sin bt \, dt = \left[ -\dfrac{b \cos bt}{e^{st}} - \dfrac{s \sin bt}{e^{st}} \right]_0^\infty$
$\displaystyle (s^2 + b^2)\int_0^\infty e^{-st} \sin bt \, dt = \left[ \dfrac{-b \cos bt - s \sin bt}{e^{st}} \right]_0^\infty$
$\displaystyle (s^2 + b^2)\int_0^\infty e^{-st} \sin bt \, dt = \left[ \dfrac{-b \cos \infty - s \sin \infty}{e^\infty} - \dfrac{-b \cos 0 - s \sin 0}{e^0} \right]$
$\displaystyle (s^2 + b^2)\int_0^\infty e^{-st} \sin bt \, dt = \left[ 0 - \dfrac{-b - 0}{1} \right]$
$\displaystyle (s^2 + b^2)\int_0^\infty e^{-st} \sin bt \, dt = b$
Thus,
$\displaystyle \int_0^\infty e^{-st} \sin bt \, dt = \dfrac{b}{s^2 + b^2}$
Therefore,
$\mathcal{L} (\sin bt) = \dfrac{b}{s^2 + b^2}$ answer
