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I have no idea how to integrate by parts

  1. Feb 2, 2005 #1
    I dont think anyone in class understands it, we went over it so quick. The only thing I seem to get is that uv - (integral) vdu = (integral) udv. You are supposed to assign u, v, dv and du, but how do you know which is u and which is v? What is the difference between du and dv? Are you integrating the variable in front of du and integrating the unit in front of dv? How do you know which is which? Why does dv sometimes have one of the variables in it and sometimes another?

    Beyond that there seem to be so many exceptions and substitutions that I can't begin to get a good handle on it. Is there a website that has basic info for lost people like me?
     
  2. jcsd
  3. Feb 2, 2005 #2
  4. Feb 2, 2005 #3
    suppose you have a function [itex]f(x) = uv[/itex]
    it's derivative is given by the product rule:

    [itex]\frac{d}{dx}[uv] = u'v + uv'[/itex]
    now if you rearrange that to
    [itex]u'v = \frac{d}{dx}[uv] - uv'[/itex]
    and then take the integral in respect to x of both sides
    [itex]\int u'v dx= \int \frac{d}{dx}[uv] dx - \int uv' dx[/itex]
    you are left with the equation for integration by parts:
    [itex]\int u'v = uv - \int uv'[/itex]

    so for a given function of two factors, you can choose either of them to be u' and v... It doesn't matter...

    Though generally you want to make u something simple to integrate and v something that becomes simpler after differentiation.
    For example [itex]\int \ln(x) dx[/itex], you should pick u' = 1 and v = ln, had you reversed those selections, you would be left with the exact same problem... meaning you still have to integrate ln(x).

    The only "exception" to that rule of thumb, is that sometimes the integral will repeat on the other side:

    [itex]\int f(x) dx = g(x) - \int f(x) dx [/itex]
    In which case you add the righ integral to the left side and divide by the coefficient (in this case 2).
     
    Last edited: Feb 2, 2005
  5. Feb 3, 2005 #4

    Galileo

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    If you forget the formula, just know that it can be derived from the product rule.
    Sometimes, it's even easier to do it directly like:

    [tex]\int f(x)g'(x)dx=\int [\frac{d}{dx}f(x)g(x)-f'(x)g(x)]dx[/tex]
     
  6. Feb 3, 2005 #5

    Hurkyl

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    Just do a lot of practice problems.

    Also there usually aren't many ways to choose u and v -- if you don't know, try them all!
     
  7. Feb 3, 2005 #6

    dextercioby

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    Hurkyl's right,there aren't many possible combination of chosing u & dv.The problem appears in the moment when u tried them all and none goes to a result...
    For example:
    [tex] \int \frac{\sin x}{x} dx [/tex]
    for 2 possible combinations and
    [tex] \int \frac{e^{2x}}{x+6}\tan x \ dx [/tex]
    for a bunch of 6 combinations...

    Daniel.
     
  8. Feb 8, 2005 #7
    is the uv - (integral) vdu method the same as the reduction formula? If not then what is the reduction formula?

    What is the end goal of integration by parts? Is it to find the area bounded by the two functions? How do you know when you've simplified it?
     
  9. Feb 8, 2005 #8

    dextercioby

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    That reduction formula (i've never heard this expression b4) is it somehow related to recurrent integrals...?Like sequences of integrals..?

    The "end goal" is to find the antiderivatives among "elementary" functions...

    Daniel.
     
  10. Feb 8, 2005 #9

    arildno

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    That hopeless u,v, du,dv mysticism..:yuck:
    It has never taught any students anything, I strongly advise you to read Galileo's post.
     
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