Laplace transform of $\sin(wt)\cdot u(t-t_0)$

Question

I need to find the laplace transform of $\sin(wt)\cdot u(t-t_0)$

The answer is $$\frac{e^{-st_0}}{\sqrt{s^2+w^2}}\cdot \sin\left(wt_0 + \tan^{-1}\frac{w}{s}\right)$$

I tried using the general definition:

$$\int_{-\infty}^{\infty}\sin(wt)\cdot u(t-t_0)e^{-st}\mathrm{d}t=\int_{t_0}^{\infty}\sin(wt)e^{-st}\mathrm{d}t$$

Edit : (After seeing @cesareo's final answer, I have corrected minor mathematical mistakes and got the same solution but I still require the above form)

After using: $$\int e^{ax}\sin(bx)\,dx=\frac{e^{ax}}{a^2+b^2}\left(a\sin(bx)-b\cos(bx)\right)$$

I get:

$$=\frac{e^{-st}}{s^2+w^2}(-s\sin(wt)-w\cos(-st))\Biggr|_{t_0}^{\infty}=\frac{e^{-st_0}}{s^2+w^2}(s\sin(wt_0)+w\cos(-st_0))$$

I am stuck here. I think I will require some trigonometrical reduction but I am not able to figure it out.

Edit 2:

I found the final solution by simply using inverse identities:

$$\frac{e^{-st_0}}{s^2+w^2}(s\sin(wt_0)+w\cos(-st_0))=\frac{e^{-st_0}}{\sqrt{s^2+w^2}}\left(\frac{s}{\sqrt{s^2+w^2}}\sin(wt_0)+\frac{w}{\sqrt{s^2+w^2}}\cos(st_0)\right)$$

Using: $$\tan^{-1}(\frac ws)=\cos^{-1}\left(\frac{s}{\sqrt{s^2+w^2}}\right)=\sin^{-1}\left(\frac{w}{\sqrt{s^2+w^2}}\right)$$

We can get the final answer in terms of arctan.

Answer 1

$$ \mathcal{L}\left(\sin(\omega t)u(t-t_0)\right)=\int_0^{\infty}e^{-s t}\sin(\omega t)u(t-t_0)dt $$

now making the change of variable $\tau = t-t_0$ we have

$$ \mathcal{L}\left(\sin(\omega t)u(t-t_0)\right)=e^{-s t_0}\int_{-t_0}^{\infty}e^{-s\tau}\sin(\omega(\tau +t_0))u(\tau)d\tau = e^{-s t_0}\int_0^{\infty}e^{-s\tau}\sin(\omega(\tau +t_0))d\tau $$

and finally as $\sin(a+b) = \sin a\cos b+\cos a\sin b$

$$ \mathcal{L}\left(\sin(\omega t)u(t-t_0)\right)=e^{-s t_0}\frac{s \sin (\omega t_0)+\omega \cos (\omega t_0)}{\omega ^2+s^2} $$

NOTE

$$ \frac{s \sin (\omega t_0)+\omega \cos (\omega t_0)}{\sqrt{\omega ^2+s^2}} = \sin\left(\omega t_0+\arctan\left(\frac{\omega}{s}\right)\right) $$

so the answer can be written also as

$$ \mathcal{L}\left(\sin(\omega t)u(t-t_0)\right)=\frac{e^{-s t_0}}{\sqrt{\omega^2+s^2}}\sin\left(\omega t_0+\arctan\left(\frac{\omega}{s}\right)\right) $$