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Problems with integration by parts

  1. Aug 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Hi all! I seem to be having trouble doing integration by parts. I seem to have a pretty clear picture of the steps I need to do but something seems to always trick me. Usually I would ask my prof but she is away for a week.

    I use the formula: uv - ∫vdu = given integration

    2. Relevant equations

    given integration = uv - ∫vdu

    3. The attempt at a solution

    Problem 1. Integrate! ∫(x^2) cosmx dx
    u = x^2 -> du = 2x dx
    dv = cosmxdt -> v = (1/m)sinmx

    Use : uv - ∫vdu
    (x^2) (1/m)sinmx - ∫(1/m)(sin(mx)) 2x dx
    I pull the constants out
    (x^2) (1/m)sinmx - (2 /m) ∫ (sin(mx)x) dx
    I’m not sure what to do next… can I use the integration by parts again? Maybe using substitution?

    Problem 2. Integrate! ∫t(sec(2t)^2) dt
    U = t -> du = 1*dt
    Dv = (sec(2t))^2 dt -> v = (½)(tan2t)
    Use : uv - ∫vdu
    t*(½)(tan2t) - ∫(½)(tan2t)dt
    simplify a little
    t/2(tan2t) – 1/2∫(tan2t)dt
    I’m not sure what to do with the last term…. not sure how to integrate tan2t

    Problem 3. Integrate! ∫(e^2x)sin3x dx
    Not sure what to choose for u… usually I pick something that will simplify when I take the derivative. In this case (e^2x) does not simplify anything and sin3x just becomes 3cos3x which is more complex.

    Any help would be appreciated!

    Tjvelcro
     
  2. jcsd
  3. Aug 7, 2010 #2
    Yeah use by parts for get rid of the x in ∫ (sin(mx)x)dx then simply integrate cos(mx).

    ∫(tan2t)dt is something you would find on an integral table, but I'm not a fan of them.

    tanx = sinx / cos x

    u = sinx
    du = cosx dx

    Remember cos^2(x) = 1- sin^2(x)


    This kind of integral usually just flips around and back since derievative of e^x and sinx still involve e^x and cosx and sinx respectively.

    Set I = ∫(e^2x)sin3x dx

    ∫(e^2x)sin3x dx = (1/2)(e^2x)sin3x dx + 3/2 ∫(e^2x)cos3x dx

    We can do the some thing with ∫(e^2x)cos3x dx and we will end up with our original integrand of ∫(e^2x)sin3x dx
     
  4. Aug 7, 2010 #3

    Char. Limit

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    That would be extremely difficult, considering you'd end with du in the denominator...

    Try the substitution...

    u=cos(x)
    du = - sin(x) dx

    instead. It will work out far better, and you won't need any more trig identities, except perhaps...

    [tex]sec(x) = \frac{1}{cos(x)}[/tex]
     
  5. Aug 7, 2010 #4
    What ?
    ∫(tan2t)dt

    tanx = sinx / cos x

    ∫tanxdx

    u = sinx
    du = cosx dx
    du/cosx =dx

    ∫u/(1-u^2) du -- This is a simple patial fractions.
     
  6. Aug 7, 2010 #5

    Char. Limit

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    And my substitution gives a simple -1/u du. Integrate to -ln(u)+C and unsubstitute. No partial fractions at all.
     
  7. Aug 7, 2010 #6
    Well great, you have an even quicker solution ! I was just a bit puzzled when you said my suggestion would be extremely difficult.
     
  8. Aug 7, 2010 #7

    Char. Limit

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    Gold Member

    I neglected to consider the trig identity you used. TBH, I actually automatically go from [itex]\int tan(u) du[/itex] to the solution.
     
  9. Aug 7, 2010 #8
    Of course, everyone has a different way of approaching a problem. Heck, if you asked me on a different day ,this same question , I may provide a different substitution/solution .

    I have to admit your solution is more enticing.
     
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