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Homework Help: Integration by parts question

  1. Jun 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Gosh I've been asking a lot of questions lately...anyways...

    I tried this question two separate times and couldn't manage to figure out where i went wrong...

    [tex]\int e^{-x}cos2x dx[/tex]

    2. Relevant equations

    [tex] uv - \int v du = \int u dvdx

    3. The attempt at a solution

    let [itex]u = e^{-x}[/itex] and [itex]dv = cos2x dx[/itex]
    so [itex]du = -e^{-x}[/itex] and [itex]v = \frac{1}{2}sin2x[/itex]

    [tex]\int e^{-x}cos2x dx = \frac{1}{2}e^{-x}sin2x + \frac{1}{2}\int e^{-x}sin2x dx[/tex]

    use integration by parts again, this time let [itex]u = e^{-x}[/itex] and [itex]dv = sin2x dx[/itex] so [itex]du = -e^{-x}[/itex] and [itex]v = \frac{-1}{2}cos2x[/itex]

    so now we have:

    [tex]\int e^{-x}cos2x dx = \frac{1}{2}e^{-x}sin2x - \frac{1}{2}e^{-x}cos2x - \frac{1}{2}\int e^{-x}cos2x dx[/tex]

    bring the [itex]\frac{1}{2}\int e^{-x}cos2x dx[/itex] to the LHS and solve for the integral.

    [tex]\frac{3}{2}\int e^{-x}cos2x dx = \frac{1}{2}e^{-x}sin2x - \frac{1}{2}e^{-x}cos2x[/tex]

    so finally:

    [tex]\int e^{-x}cos2x dx = \frac{1}{3}e^{-x}(sin2x - cos2x)[/tex]

    which is incorrect... at which step did I go wrong?

    thank you all in advance!
    Last edited: Jun 3, 2010
  2. jcsd
  3. Jun 3, 2010 #2


    User Avatar
    Homework Helper

    You forgot a factor 1/2 the second time you used integration by parts.
  4. Jun 3, 2010 #3
    ahh, in which case i would get

    \int e^{-x}cos2x dx = \frac{1}{5}e^{-x}(2sin2x - cos2x)
  5. Jun 3, 2010 #4
    thanks for pointing out my silly mistake...I have one more question that i will post up in a few minutes, this assignment is really killing me. Thanks again cyosis.
  6. Jun 3, 2010 #5


    Staff: Mentor

    Plus the constant of integration.
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