Laplace Transform for solve ODE (RLC circuit)

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I have an RLC circuit and I want to know the charge on the capacitor $q(t)$ using Laplace transform:

The diferential equation is:

$$ Lq'' + Rq' + \frac{1}{C}q = E(t),$$

where

$L = 1H , R = 20 \Omega, C = 0,005 F, q(0) = 0 e q'(0) = 0$.

$$ E(t) =\begin{cases}sin(t), & 0 \leq t \leq 2\pi\\0, & t \geq 2\pi. \end{cases} $$

Well, here is my attempt:

The Laplace transform of the diferential equation is:

$$\Big[ Ls^2 + Rs + \frac{1}{C}\Big]Q(s) = \mathcal{L}[E(t)] = E(s).$$

The inverse Laplace transform of $E(t)$ is:

$$E(t) = sin(t) - u_{2\pi}(t)sin(t)$$ which then gives: $$E(s) = \frac{1}{s^2 + 1}e^{-2\pi s}.$$

Making $ Ls^2 + Rs + \frac{1}{C}=E(s)$, we get:

$$Q(s) = \frac{1}{(s^2 + 1)(Ls^2 + Rs + \frac{1}{C})} - \frac{e^{-2\pi s}}{(s^2 + 1)(Ls^2 + Rs + \frac{1}{C})}.$$

I'm have some problem solving this Laplace inverse transform...I think I'm not solving the partial fractions well. I'm right in this point. Please how can I resolve the partial fraction and/or Inverse Laplace transform of this equation?