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#1 Re: Help Me ! » Laplace Transform of sin^2t » 2012-10-25 10:18:26

to be clear I am talking about (sin(t))^2

#2 Re: Help Me ! » Laplace Transform of sin^2t » 2012-10-25 10:17:22

Hello,

I get that the laplace transform of sin^2t = -(sin^2te^-st)/s + 2/s^3+4s evaluated from 0...infinity. 

when I evaluate the limit from 0..infinity I get that the transform to equal 0.  Did I evaluate that right?

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