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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?

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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.

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  • $\begingroup$ i find out my mistake now, thanks a lot! =) $\endgroup$ – user118035 Jan 17 '14 at 11:06
  • $\begingroup$ You are welcome. Please recall to upvote and/or accept answers that are helpful. Regards $\endgroup$ – Amzoti Jan 17 '14 at 13:19
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Consider $$\mathcal{L}[a\cos bx](s)=a\int_0^\infty e^{-sx}\cos(bx)\mathrm dx$$ Then recall that $\cos x=\text{Re}\,e^{ix}$. Hence we have $$\mathcal{L}[a\cos bx](s)=a\,\text{Re}\int_0^\infty e^{-(s-ib)x}\mathrm dx$$ Which is $$\begin{align} \mathcal{L}[a\cos bx](s)&=a\,\text{Re}\,\frac1{s-ib}\\ &=a\,\text{Re}\,\frac{s+ib}{s^2+b^2}\\ &=\frac{as}{s^2+b^2} \end{align}$$ Your case is of course given by $a=2$ and $b=3$.

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