Determine the laplace transform of the function $f(t)=2\cos(3t)$, without using the table of Laplace transforms.
I use by part integration to solve it, with $u=e^{-st},\, du/dt=-se^{-st}$ and $v=\frac{2}{3}\sin(3t)$. I get $0$ at the end since $\lim_{t\rightarrow \infty} e^{-st}$ and $-se^{-st}$ is $0$. However, the graphic calculator shows the answer is $1/5$.
So how to solve this question?
We do integration by parts twice and arrive at:
$$\begin{align} \mathscr{L} (2 \cos(3t)) & = \int_0^{\infty} ~ 2 \cos(3 t) e^{-s t}~dt \\ \\ &= \dfrac{e^{-s t}( 6 \sin(3 t) - 2 s \cos(3 t))}{s^2 + 9}~\Bigr|_{t=0}^{t = \infty} \\ \\ &= \dfrac{2s}{s^2+9}\end{align}$$
An alternate approach is to utilizes Euler's formula.