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Indefinite Integral with integration by parts

  1. May 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Evaluate ∫ecos2θ dθ

    2. Relevant equations
    Integration by parts formula
    ∫udv = uv -∫vdu

    3. The attempt at a solution
    So in calc II we just started integration by parts and I'm doing one of the assignment problems. I know I need to do the integration by parts twice, but I've hit a loop, or so it seems, which I know isn't right. Maybe someone can see my faults?

    I set:
    u = cos(2θ) v = -e
    du = -2sin(2θ) dθ dv = e

    So by the formula I got:
    = [cos(2θ)] [-e] -2∫ [-e] [sin(2θ) dθ]

    Here I used the second integration by parts:

    u = sin2θ v = e
    du = 2cos(2θ) dθ dv = -e

    Solving by the formula again:
    = [sin(2θ)] [e] -2∫ [e] [cos(2θ) dθ]

    I'm not too sure where I've made my algebraic error, or if I'm on the right track and this won't just put my into a loop giving the same equation above? First question I've posted, so hopefully it follows the format ok, if it doesn't, chime in and let me know so I can fix it properly. Thanks in advance everyone!
     
  2. jcsd
  3. May 16, 2015 #2

    phion

    User Avatar
    Gold Member

    Your substitutions are correct, but you're stopping a little short. The trick with integration by parts composed of an exponential function and a trigonometric function is to add to both sides the first integral you find that is a multiple of your original integral. From there, it should be obvious where to go next.
     
  4. May 16, 2015 #3
    Ohh, hat's definitely what I'm missing! Thanks a bunch for the help!
     
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