I am not certain how to begin this problem. Someone please point me in the right direction.
Problem
Using the two given formulas ($1$ and $2$ below) show that:
$$L^{-1}\{\frac{1}{(s^2+\omega^2)^2}\}=\frac{1}{2\omega^3}(\sin(\omega t)-\omega t\cos(\omega t))$$
The given formulas:
$$L_1\{t\cos(\omega t)\}=\frac{s^2-\omega^2}{(s^2+\omega^2)^2}, L_2\{t\sin(\omega t)\}=\frac{2\omega s}{(s^2+\omega^2)^2} $$
We write $$ \begin{align} \frac{1}{(s^2+\omega^2)^2}&=\frac{1}{2\omega^2}\frac{2\omega^2}{(s^2+\omega^2)^2} =\frac{1}{2\omega^2}\Bigl[\frac{\omega^2-s^2+\omega^2+s^2}{(s^2+\omega^2)^2}\Bigr]\\ &=-\frac{1}{2\omega^2}\frac{s^2-\omega^2}{(s^2+\omega^2)^2}+\frac{1}{2\omega^2}\frac{1}{s^2+\omega^2}\\ &=-\frac{1}{2\omega^2}\frac{s^2-\omega^2}{(s^2+\omega^2)^2}+\frac{1}{2\omega^3}\frac{\omega}{s^2+\omega^2}\\ \end{align} $$ The first part is the Laplace transform of $-\frac{1}{2\omega^2}t\cos(\omega t)$ according to your first formula.
Differentiating the second part (with respect to $s$), $$ \frac{d}{ds}\frac{1}{2\omega^3}\frac{\omega}{s^2+\omega^2} = -\frac{1}{2\omega^3}\frac{2\omega s}{(s^2+\omega^2)^2} $$ Using that $\mathcal{L}(tf(t))=-F'(s)$ we find, using your second formula, that the second part is the Laplace transform of $$ \frac{1}{2\omega^3}\sin(\omega t). $$ Adding, we get $$ \mathcal{L}^{-1}\{\frac{1}{(s^2+\omega^2)^2}\}=\frac{1}{2\omega^3}(\sin(\omega t)-\omega t\cos(\omega t)) $$