Question

laplace transform of sin^3 2t

Answer 1

Here are some basic transforms you can use,

http://image.mathcaptain.com/cms/images/83/laplace-tranform-formulas-list.png

Answer 2

thanks ,tried using them,but the main problem is coming from the power rised(3)

Answer 3

\[\large \mathscr{L}\left[\sin^3(2t)\right] \qquad = \qquad
\mathscr{L}\left[\sin(2t)\sin^2(2t)\right] \qquad =\qquad
\\ \large \mathscr{L}\left[\sin(2t)\left(1-\cos^2(2t)\right)\right] \qquad = \qquad \mathscr{L}\left[\sin(2t)\right]-\mathscr{L}\left[\sin(2t)\cos^2(2t)\right]\]

Answer 4

Woops it got cut off..\[\large \mathscr{L}\left[\sin(2t)\right]-\mathscr{L}\left[\sin(2t)\cos^2(2t)\right]\]

That first term isn't too bad right?
For the other one, use the definition of the Laplace Transform.
You'll have to do integration by parts a couple times.

Answer 5

\[L\{sin^3(2t)\}=\int_{0}^{inf}e^{-st}sin^3(2t)~dt\]
which still amounts to some by parts stuff

Answer 6

I propose doing:
\[\mathcal{L} \left( \sin^3(2t) \right)=\int\limits_0^{\infty}e^{-st} \left( \frac{e^{2ti}-e^{-2ti}}{2i}\right)^3dt\]
This **should** make your life easier.