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Trig derivatives

  1. Jan 15, 2008 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    integration by parts formula

    3. The attempt at a solution

    = 9/10e[tex]^{2x}[/tex](sin(3x)-1/3cos(3x))

    I don't think I am taking the derivative of sin(3x) correctly

    thanks for taking a look :)
  2. jcsd
  3. Jan 15, 2008 #2


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    [tex]\int e^{2x}sin(3x)dx[/tex]

    u=e^2x so that du=2e^2x dx

    dv=sin(3x)dx ; v=-1/3 cos(3x)

    but either way you would have to do integration by parts twice so I am thinking that is not the answer(BUT I can be wrong as I did not really work it out)
  4. Jan 15, 2008 #3
    Hint :

    What you do is you integrate it by parts two times.

    A multiple of the original integral will appear in the answer.
  5. Jan 15, 2008 #4
    I tried it with u = e^2x ; dv = sin3xdx

    I think i got the multiple and im stuck here:

    = -1/3e[tex]^{2x}cos(3x) + 4/9[/tex][tex]\int[/tex]sin(3x)e[tex]^{2x}[/tex]
  6. Jan 15, 2008 #5


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    you are supposed to get
    [tex]\int e^{2x}sin(3x)dx=\frac{-e^{2x}}{3}cos(3x)+ \frac{2}{3}\int e^{2x}cos(3x)dx[/tex]

    then do integration by parts again
  7. Jan 15, 2008 #6
    I got it, thanks guys

    1/13 e^2x (2sin3x - 3cos3x) = integral
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