Determine the Laplace Transform for the following function:
$$ f(t) = t^2\cos(2t + 30^\circ) $$ We begin by reworking the expression using the following trigonometric identity for sum/difference:
$$ \cos(A \pm B) = \cos(A) \cos(B) \mp \sin(A) \sin(B) $$
Proceeding onward... $$ \begin{align} f(t) & = t^2\cos(2t + 30^\circ) \\ & = t^2[\cos(2t)\cos(30^\circ) - \sin(2t)\sin(30^\circ)] \\ & = t^2 \Big[ \frac{\sqrt{3}}{2} \cos(2t) - \frac{1}{2} \sin(2t) \Big] \\ f(t) & = \frac{\sqrt{3}}{2}t^2 \cos(2t) - \frac{1}{2} t^2\sin(2t) \qquad,(Eqn\;1) \end{align} $$ We now proceed to determine the Laplace transform of equation #1: $$ \begin{align} \mathcal{L}[f(t)] &= F(s) \\ &= \frac{\sqrt{3}}{2}\mathcal{L}[t^2 \cos(2t)] - \frac{1}{2} \mathcal{L}[t^2\sin(2t)] \qquad,(Eqn\;2) \end{align} $$
Recall the definition of the Frequency Differentiation Property. This is also property #30 in our Laplace Transform table... $$ \mathcal{L}[t^nf_1(t)] = F(s) = (-1)^n \frac{d^nF_1(s)}{ds^n} $$ ...and apply it to the terms on both sides of the minus sign in equation #2:
$$ \begin{align} F(s) &= \frac{\sqrt{3}}{2} \Big[ (-1)^2 \frac{d^2}{ds^2}\Big(\mathcal{L}[\cos(2t)]\Big) \Big] - \frac{1}{2}\Big[ (-1)^2 \frac{d^2}{ds^2} \Big(\mathcal{L}[\sin(2t)] \Big)\Big] \\ &= \frac{\sqrt{3}}{2} \Big( \frac{d^2}{ds^2}\Big[ \frac{s}{s^2+4} \Big] \Big) - \frac{1}{2} \Big( \frac{d^2}{ds^2} \Big[ \frac{2}{s^2+4} \Big] \Big) \\ &= \frac{\sqrt{3}}{2} \Big( \frac{d}{ds}\Big[ \frac{(s^2+4)-[s(2s)]}{(s^2+4)^2} \Big] \Big) - \frac{1}{2}\Big( \frac{d}{ds}\Big[ frac{0-2(2s)}{(s^2+4)^2} \Big] \Big) \\ &= \frac{\sqrt{3}}{2} \Big( \frac{d}{ds}\Big[ \frac{4-s^2}{(s^2+4)^2}\Big]\Big) - \frac{1}{2}\Big( \frac{d}{ds}\Big[ \frac{-4s}{(s^2+4)^2} \Big] \Big) \end{align} $$
$$ \begin{align} &=\frac{\sqrt{3}}{2}\Big[ \frac{(s^2+4)^2(-2s)-(4-s^2)(2)(2s)(s^2+4)}{(s^2+4)^4}\Big] - \frac{1}{2}\Big[\frac{(s^2+4)^2(-4)+4s(2)(2s)(s^2+4)}{(s^2+4)^4}\Big] \\ &= \frac{\sqrt{3}}{2} \cancel{(s^2+4)}\Big[ \frac{(-2s)(s^2+4)-4s(4-s^2}{(s^2+4)^{\cancel{4}3}}\Big] - \frac{1}{2}\cancel{(s^2+4)}\Big[ \frac{{-4)(s^2+4)+16s^2}}{(s^2+4)^{\cancel{4}3}} \Big] \\ \end{align} $$
$$ \begin{align} &= \frac{\sqrt{3}}{2}\Big[ \frac{-2s^3-8s-16s+4s^3}{(s^2+4)^3} \Big] - \frac{1}{2}\Big[ \frac{-4s^2+16s^2-16}{(s^2+4)^3} \Big] \\ &= \sqrt{3}\Big[ \frac{s^3-12s}{(s^2+4)^3} \Big] - \Big[ \frac{6s^2-8}{(s^2+4)^3} \Big] \end{align} $$
$$ F(s) = \frac{s^3 \sqrt{3} - 12s\sqrt{3} - 6s^2 + 8}{(s^2+4)^3} $$
Continue on to Properties of the Laplace Transform example problem #4...