Find the Laplace transform of integral(from 0 to x) sin(2t) dt

Question

Find the Laplace transform of $\int_0^x\,\sin\,(2t)\,dt$

So basically, $$\int_0^x\,\sin\,(2t)\,dt = -\frac{1}{2}(\cos\,(2x) - 1)$$

So $$\mathcal{L}\{\cos\,(2x)\} = \dfrac{s}{s^2 + 4}$$

$\mathcal{L}\{-1\} = \mathcal{L}\{-\delta(x)\} = -1$

So I got the answer to be $1 - \dfrac{s}{s^2+4}$

The answer says, $\dfrac{2}{s^3+4s}$. Any ideas?

Answer 1

Laplace transform of a constant is

$$\mathcal L\{c\} = \dfrac{c}{s}$$

So

$$\mathcal L\{\dfrac{1}{2}(1-\cos(2x))\} = \dfrac{1}{2}\left(\dfrac{1}{s} - \dfrac{s}{s^2+4}\right) = \dfrac{1}{2}\dfrac{s^2+4 - s^2}{s(s^2+4)} = \dfrac{2}{s^3+4s}$$