Laplace tranform of unitstep (heaviside) function

[dfang]

Homework Statement

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I know for t[u(t)-u(t-2)], we can simplify that to tu(t)-((t-2)+2)u(t-2) which then gives us tu(t)-(t-2)u(t-2)-u(t-2). Now, the laplace transform seems trivial but I am having problems with this equation:
sin(t)[u(t)-u(t-pi)]

Homework Equations

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3. The Attempt at a Solution
sin(t)[u(t)-u(t-pi)]
=sin(t)u(t) - sin(t)u(t-pi)
= sin(t)u(t) - sin ((t-pi)+pi)u(t-pi)
Where do we go from here as the t-pi+pi is inside the sine i.e its a function of sine. We can't exactly take it out and multiply over. How can we take this to the form where I can take the laplace transform easily ?
Similar problem where I am having trouble: e^-t u(t-2). This one I do not even know where to start as the -t is in the exponent this time. Thanks.

 

[LCKurtz]

Try using this: $$\mathcal{L}f(t)u(t-a)=\int_a^\infty e^{-st}f(t)~dt=\int_0^\infty e^{-s(v+a)}f(v+a)~dv$$ $$ =
e^{-as}\int_0^\infty e^{-sv}f(v+a)~dv =e^{-as}\int_0^\infty e^{-st}f(t+a)~dt=e^{-as}\mathcal{L}f(t+a) $$