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    Ok, so the goal is to find the Laplace transform of f(t)=cos(at) where a is a real constant.

    L=\displaystyle\int^\infty_0 e^{-st}cos(at)\ dt

    Working it out using integration by parts twice I arrive at
    \displaystyle\lim_{t\to \infty} \frac{e^{-st}cos(at)}{s}-\frac{a}{s}[\frac{e^{-st}sin(at)}{s}+\frac{a}{s}\displ  aystyle\int^t_0 e^{-st}cos(at) dt]

    At this point I would normally substitute the remaining integral for what's on the other side of the original equation but here there is nothing. Is there some trick I'm not seeing or should I just give up and use a table of integrals?
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    What's happened to the t=0 limit when you did your first integration by parts?
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    Cos (at) = the real part of e^iat
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    (Original post by D-Day)
    Ok, so the goal is to find the Laplace transform of f(t)=cos(at) where a is a real constant.

    L=\displaystyle\int^\infty_0 e^{-st}cos(at)\ dt

    Working it out using integration by parts twice I arrive at
    \displaystyle\lim_{t\to \infty} \frac{e^{-st}cos(at)}{s}-\frac{a}{s}[\frac{e^{-st}sin(at)}{s}+\frac{a}{s}\displ  aystyle\int^t_0 e^{-st}cos(at) dt]

    At this point I would normally substitute the remaining integral for what's on the other side of the original equation but here there is nothing. Is there some trick I'm not seeing or should I just give up and use a table of integrals?
    If you let your integral equal to I, then you'll have I = "some stuff with an I in". You then solve for I.

    In this case you'll have:

    I = 1/s - (a^2/s^2) * I.

    Sorry your solutions may or may not be correct (I've quoted you and it has gone into latex code) but as long as you check your integration by parts to make sure you've included all your limits etc you'll end up with:

    I = s/(s^2 + a^2) where I is the Laplace of cos(at).

    If you can't get that I'll check your integration by parts with mine.
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    Another way to integrate this is to notice that it's the real part of e^st*e^iat.
 
 
 
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