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Partial Fractions prob

  1. Dec 2, 2008 #1
    [tex]\int e^{ax}cosbx[/tex]

    This one is driving me insane.

    So I used e^ax as u and cosbx dx as dv. And then I did it again using e^ax as u and sinbx as dv which left me with [tex]\int e^{ax}cosbx = \frac{1}{b}e^{ax}sinbx + \frac{a}{b^{2}}e^{ax}cosbx - \frac{a^{2}}{b^{2}}\int e^{ax}cosbxdx[/tex]

    I have no ideas if this is right and no clue what do after if it is. Thanks.
     
  2. jcsd
  3. Dec 2, 2008 #2
    Notice that you get back the same integral on the right side, so move it to the left and factor.

    Then check your answer by differentiation.
     
  4. Dec 2, 2008 #3
    ok this is probably a stupid algebra question but when I move the integral from the right side to the left side then how do I get rid of the a^2/b^2 that's stuck to it?
     
  5. Dec 2, 2008 #4
    nvm (word count)
     
  6. Dec 3, 2008 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    And, finally, this problem had nothing at all to do with "partial fractions".
     
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