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Trivial integration by parts?

  1. Sep 24, 2006 #1
    Why is it that when I do integration by parts on cyclic functions such as (sinx)e^(inx), I get a trivial answer like C=C, C is a constant

    Have I done something wrong or are there other methods of doing those integrals?
    Last edited: Sep 24, 2006
  2. jcsd
  3. Sep 24, 2006 #2
    I don't understand your question, could you give a better example of what you mean? Integrals like this usually work out nicely by using parts twice and then collecting like terms of the original integral on one side of teh equation (I'm not sure how much sense that made if any).
  4. Sep 24, 2006 #3
    As soon as you get the integral of your original function, stop using parts. Merge it with the original function on the left and divide by the new constant. Using parts again just proves that it is equal (c=c)
  5. Sep 25, 2006 #4
    I did the integral originally posted with a different order, exponetial on the left and it worked out nicely. The first time I did it, I had sin on the left and it turned into the trivial C=C (following all the usual rules). Looks like the order of integration matters.
  6. Sep 25, 2006 #5


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    If, integrating something like [itex]\int f(x)g(x)dx[/itex], you do an integration by parts, letting u= f(x), dv= g(x)dx, you will, of course, get another integral [itex]\int vdu[/itex]. If you do another integration by parts, this time reversing the roles (letting u= the old v and dv= the old u dx) you will just reverse what you did in the first integration and, yes, everything will cancel.
  7. Sep 26, 2006 #6
    I see what you are getting at here, I realised that I made the exact mistake you pointed out here.
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