Laplace transform of $-{e}^{bt}\mathrm{sin}({\omega}_{0}t)u(-t)$

I know that Laplace transform of

$${e}^{-\alpha t}\mathrm{sin}({\omega}_{0}t)u(t)$$

is $\frac{{\omega}_{0}}{(s+\alpha {)}^{2}+{\omega}_{0}^{2}}$ where $Re(s)>-Re(\alpha )$ and $u(t)={\mathbf{1}}_{t\ge 0}$. Knowing that, how can I find the Laplace transform of$$-{e}^{bt}\mathrm{sin}({\omega}_{0}t)u(-t)\text{}\text{}?$$

I know that Laplace transform of

$${e}^{-\alpha t}\mathrm{sin}({\omega}_{0}t)u(t)$$

is $\frac{{\omega}_{0}}{(s+\alpha {)}^{2}+{\omega}_{0}^{2}}$ where $Re(s)>-Re(\alpha )$ and $u(t)={\mathbf{1}}_{t\ge 0}$. Knowing that, how can I find the Laplace transform of$$-{e}^{bt}\mathrm{sin}({\omega}_{0}t)u(-t)\text{}\text{}?$$