Laplace transform of $f(t)=t^{2}\cos(t)-t\cos^{2}(t)$.

Question

My question is: find the Laplace transform of this function, $f(t)=t^{2}\cos(t)-t\cos^{2}(t)$. First I remove $\cos^{2}t$, $\cos^{2}t=\frac{1}{2}+\frac{1}{2}\cos(2t)$. Then for $t^{2}\cos(t)$, I need to manipulate it to become something else. How to proceed?

Answer 1

$ \newcommand{\L}[1]{\mathcal{L}\big\{#1\big\}(s) } $ You propose that

$$f(t) = t^2 \cos(t) - \frac t 2 - \frac 1 2 t \cos(2t)$$

We then may make use of the following identities: when $n \in \Bbb Z^+$ and $\alpha,\beta \in \Bbb R$

\begin{align*} \L{\alpha f(t) + \beta g(t)} &= \alpha \cdot \L{f(t)} + \beta \cdot \L{g(t)} \tag{linearity} \\ \L{t^n} &= \frac{n!}{s^{n+1}} \tag{3}\\ \L{t \cos(\alpha t)} &= \frac{s^2 - \alpha^2}{\left( s^2 + \alpha^2 \right)^2} \tag{10}\\ \L{t^n g(t)} &= (-1)^n \frac{\mathrm{d}^n}{\mathrm{d}s^n} \L{g(t)} \tag{30} \\ \L{\cos(\alpha t)}&= \frac{s}{s^2 + \alpha^2} \tag{8} \end{align*}

(Equations as numbered per here.)

Specifically, use linearity of $\mathcal{L}$ to split it up across the summands and pull out the relevant constants, apply $(3)$ to the middle term, $(10)$ to the third term, and $(30)$ and then $(8)$ to the first.

I'll leave the details of the calculations to you.