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Integration by Parts

  1. Dec 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral
    (e^-theta) cos(2theta)

    I got this as my answer
    e^(-theta)-sin(2theta)+cos(2theta)e^(-theta)+C
    But it was wrong
    All help is appreciated.
     
  2. jcsd
  3. Dec 4, 2009 #2

    rock.freak667

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    well you don't have "1/2" in there anywhere

    u=e so du=?

    dv= cos2θ dθ so v= ?
     
  4. Dec 4, 2009 #3
    i got du= e^-theta
    and v= sin(2theta)
     
  5. Dec 4, 2009 #4

    rock.freak667

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    v would be "1/2 sin2θ", right?
     
  6. Dec 4, 2009 #5
    This is probably a stupid question but why would it be?
     
  7. Dec 4, 2009 #6

    rock.freak667

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    [tex]\int cos n\theta d\theta[/tex]

    let t = nθ ⇒ dt = ndθ or dt/n = dθ (n is a constant)

    [tex]\therefore \int cos n\theta d\theta \equiv \int \frac{cos t}{n} dt = \frac{1}{n} sin t=\frac{1}{n}sin(n\theta)[/tex]


    see where the 1/2 comes from?
     
  8. Dec 4, 2009 #7
    ok yeah i think i got it, so now I'm at
    e^-θ - sin(2θ) - ∫1/2sin(2θ) * e^-θ
    what do i do with the ∫1/2sin(2θ) * e^-θ?
    Take the anti derivative right?
    would that be -(1/2)cos(2θ) * e^-θ?
     
  9. Dec 4, 2009 #8

    rock.freak667

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    integrate by parts again
     
  10. Dec 5, 2009 #9

    HallsofIvy

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    What is the derivative of [itex]sin(2\theta)[/itex]?

    In general, if [itex]\int f(x)dx= F(x)+ C[/itex] then to integrate [itex]\int f(ax+b) dx[/itex], let u= ax+ b so that du= a dx or (1/a)du= dx. The integral becomes [itex](1/a)\int f(u)du= (1/a)F(u)+ C= (1/a)F(ax+ b)+ C[/itex].

    That is, for f(ax+b), just as, if you were differentiating, you would have to multiply by a (by the chain rule), so when integrating, you divide by a.

    (That works for a simple linear substitution.
     
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