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When exactly does the tabular method for integration by parts fail?

  1. Feb 12, 2012 #1
    I found this interesting but different way to solve integration by parts problems on the internet: http://imageshack.us/photo/my-images/854/integration20by20parts2.jpg/

    It seems to work well for me when doing most textbook problems, except when the integrand contains a natural logarithm. I just want to know, are there other special circumstances in which you'd have to do use the definition of integration by parts to carry out the integration?

    Thanks in advance.
  2. jcsd
  3. Feb 12, 2012 #2
    You need to be able to differentiate one of the factor to 0, at which point the table ends. This fails when say, you have integrand of the form of product of exponential and cosine or sine, neither which can be differentiated to zero [there are however also clever way to deal with this kind of integrand, but I have to go now...].
  4. Feb 12, 2012 #3
    Interesting, and which method is this? I've tried searching for it and found nothing. Thanks in advance.
  5. Feb 12, 2012 #4


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    He's probably referring to integrals of the form$$\int e^x\sin x\ dx$$ or similar with a cosine. Those require two integrations by parts and then solving for the unknown integral. But an easier way is to observe the ##e^{ix}=\cos x + i \sin x## so the above integral can be done by just taking the imaginary part of $$\int e^xe^{ix}\ dx = \int e^{(1+i)x}\ dx$$Just work that using the exponential rules and rationalize it. You get$$\int e^x\sin x\ dx$$by taking the imaginary part, and for free you also get$$\int e^x\cos x\ dx$$by taking the real part.
  6. Feb 12, 2012 #5
    I actually have in mind the following:

    You can show that the following is true:

    Suppose that [itex]\dfrac{d^2f}{dx^2}=nf[/itex] and [itex]\dfrac{d^2g}{dx^2}=mg[/itex], that is if the functions are constant multiple of themselves, and [itex]m \neq n[/itex], then the integral of the product [itex]fg[/itex] is given by

    [itex]\displaystyle\int fg dx = \frac{f\frac{dg}{dx}-\frac{df}{dx}g}{m-n} + C[/itex].

    Note that there is no need to carry out any integration!
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