In order to solve the initial value problem $$ y'' - 2y' + 7y = \sin(t);\quad y(0) = 0, y'(0) = 0,$$ I need to find the inverse Laplace transform of $$ \frac{1}{ (s^2 +1)(s^2 - 2s + 7)} ,$$ this is after simplification of terms. So what I know is that I could use the convolution theorem because $$ F(s) = \frac{1}{ (s^2 +1)} \\ G(s) = \frac{1}{(s^2 - 2s + 7)} ,$$ and from here I can say that $$ (f * g)(t) = \sin(t) * \frac{e^t \sin(\sqrt{6}t)}{\sqrt{6}},$$ turning this last result into a convolution integral. So, is there a way to avoid using the convolution theorem, or else, a $\textit{simple trick}$ to solve the integral?
(NOTE: I see I could use some trigonometric identities to turn the product of sines into sums.)
Hint. By a partial fraction decomposition, one may avoid convolution, getting $$ \frac{20}{\left(s^2+1\right) \left(s^2-2 s+7\right)}=\frac{s+3}{\left(s^2+1\right)}-\frac{s+1}{\left(s^2-2 s+7\right)} $$ then, by using the Inverse Laplace Transform, we have $$ \mathcal{L}^{-1}\left(\frac{s+3}{\left(s^2+1\right)}\right)(t)=3 \sin (t)+\cos (t), $$$$ \mathcal{L}^{-1}\left(\frac{s+1}{(s-1)^2+6}\right)(t)=\frac{e^t }{\sqrt{6}}\cdot \left(2 \sin \left(\sqrt{6} t\right)+\sqrt{6} \cos \left(\sqrt{6} t\right)\right). $$