Laplace transform of $\cos(2t-(\frac\pi3))$ Problem
I need to find the Laplace transform of $\cos(2t-(\frac\pi3))$
Attempt
I've tried to look up some relevant formulae in my book, but I can't find anything that looks useful. I suspect there is something there, that I'm just not seeing.
Wolfram Alpha suggests the answer is $\displaystyle\frac{s+2\sqrt3}{2s^2+8}$, but I can't "reverse engineer" it either.
Any help appreciated!
 A: Case 1: $f(t)=\cos\left(2t-\frac{\pi}{3}\right) u\left(t\right)$
Observing that $\cos(\omega t -\phi)=\cos(\omega t)\cos\phi+\sin(\omega t)\sin\phi$ and that $\mathcal L\left\{\cos(\omega t)\right\}=\frac{s}{s^2+\omega^2}$ and $\mathcal L\left\{\sin(\omega t)\right\}=\frac{\omega}{s^2+\omega^2}$, we have
\begin{align}
\mathcal L\left\{\cos(\omega t -\phi)\right\}&=\mathcal L\left\{\cos(\omega t)\right\}\cos\phi+\mathcal L\left\{\sin(\omega t)\right\}\sin\phi\\
&=\frac{s}{s^2+\omega^2}\cos\phi+\frac{\omega}{s^2+\omega^2}\sin\phi\\
&=\frac{s\cos\phi+\omega\sin\phi}{s^2+\omega^2}
\end{align}
For $\omega=2$ and $\phi=\frac{\pi}{3}$, $\cos\left(\frac{\pi}{3}\right)=\frac 1 2 $ and $\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt 3}{2}$ we have
$$
\mathcal L\left\{\cos\left(2t-\frac{\pi}{3}\right)u\left(t\right)\right\}=
\frac{\frac{s}{2}+\sqrt{3}}{s^2+4}=\frac{s+2\sqrt{3}}{2s^2+8}
$$
Case 2: $f(t)=\cos\left(2t-\frac{\pi}{3}\right) u\left(2t-\frac{\pi}{3}\right)$
Use the Time shifting property
$$  \mathcal L\left\{f(t - a) \right\}=\mathrm e^{-as} F(s) $$
and that for $f(t)=\cos(at)$
$$
\mathcal L\left\{\cos(\omega t)\right\}=F(s)=\frac{s}{s^2+\omega^2}
$$
So $$\mathcal L\left\{\cos(\omega(t-a))\right\}=\mathrm e^{-as}\frac{s}{s^2+\omega^2} $$ and for $\omega=2$ and $a=\frac{\pi}{6}$ we have
$$
\mathcal L\left\{\cos\left(2t-\frac{\pi}{3}\right)u\left(t-\frac{\pi}{6}\right)\right\}=\mathrm e^{-\frac{\pi}{6}s}\frac{s}{s^2+4}
$$
A: You may use the basic properties of the Laplace transform, or just brute force as I am willing to do.
$$\int_{0}^{+\infty}\cos\left(2t-\frac{\pi}{3}\right)e^{-st}\,dt = \text{Re}\int_{0}^{+\infty}\exp\left((2i-s)t-\frac{\pi i}{3}\right)\,dt=\text{Re}\left(\frac{\omega}{2i-s}\right).$$
