Stuck Partway Through This Laplace Transform

Question

"Use a Laplace Transform to solve $x''+x = sin2t$; $x(0) = 0, x'(0) = 0$"

My Work So Far

$x'' = s^2X(s) - sx(0) - x'(0)$

$\therefore $

$s^2X(s) - sx(0) - x'(0) +X(s) = \frac{2}{s^2+4}$

$s^2X(s) +X(s) = \frac{2}{s^2+4}$

$X(s) =\frac{2}{(s^2+4)(s^2+1)}$

From here, though, I'm not sure what the next step would be to come up with a form that can be solved for the inverse transform. Any help moving past this step would be appreciated.

Answer 1

$$X(s) =\frac{2}{(s^2+4)(s^2+1)}=\frac{2}{-3}\left(\frac{1}{s^2+4}-\frac{1}{s^2+1}\right)$$ then $$x(t) =\frac{2}{-3}\left(\frac{1}{2}\sin2t-\sin t\right)$$

Answer 2

HINT:

$$\frac{1}{(s^2+1)(s^2+4)}=\frac{1/3}{s^2+1}-\frac{1/3}{s^2+4}$$