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Homework Help: Forgot how to integrate yes! t*cos(Pi*t)

  1. Oct 12, 2005 #1
    forgot how to integrate!! yes! t*cos(Pi*t)

    Hello everyone i'm integrating a position vector and i'm stuck on integrating the j unit. t*cos(Pi*t);
    the answer i got with maple is:
    but i have no idea how maple busted that out.
    if i let u = cos(Pi*t);
    du = sin(Pi*t)*Pi dt;
    1/Pi du = sin(Pi*t);
    but i don't see how this is helping me any...
  2. jcsd
  3. Oct 12, 2005 #2


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    Use integration by parts: f = t and dg = cos(pi*t)dt

    Then [itex]\int {fdg = fg - } \int {gdf} [/itex]
  4. Oct 12, 2005 #3
    Thanks for the responce but i'm still messing it up!
    I let f = t; dg = cos(Pi*t) dt;
    df = 1;
    i integrated dg, to get g, and got:
    g = [t*sin(Pi*t)]/Pi;

    then u said:
    fg - integral(g*df);
    (1)([t*sin(Pi*t)]/Pi) - integral (t*sin(Pi*t)]/Pi)(1); but now i'm stuck integrating this function by parts too?
  5. Oct 12, 2005 #4
    Your integral for dg has found a factor of t for some reason, your integral should be:

    \int \cos (\pi t ) dt = \frac{1}{\pi} \sin(\pi t)
  6. Oct 12, 2005 #5
    i had that, but the def says: [itex]\int {fdg = fg - } \int {gdf} [/itex] so doesn't this mean i have to take f which is t, and multiply it by g? which is [tex]
    \int \cos (\pi t ) dt = \frac{1}{\pi} \sin(\pi t)
    [/tex] thats where i got that t from
  7. Oct 12, 2005 #6
    Remember the integral on the RHS is asking for the derivative of f, so we have

    \int t\cos(\pi t) dt = \frac{t}{\pi}\sin(\pi t) - \frac{1}{\pi} \int \sin( \pi t ) dt
    Last edited: Oct 12, 2005
  8. Oct 12, 2005 #7
    ohhh thanks again sqrt!
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