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IBP Paradox for cot x integral

  1. Nov 19, 2011 #1
    Greetings! I hope this is the correct forum for my question (this is my first post, here).

    The problem statement

    Find: [itex]\displaystyle\int \cot x \ \mathrm{d}x[/itex]

    A Solution
    We can re-write this integral into a convenient form: [itex]\displaystyle\int \cot x \ \mathrm{d}x = \displaystyle\int \dfrac{\cos x}{\sin x} \ \mathrm{d}x[/itex]

    Now, let [itex]u=\sin x[/itex]. Then, [itex]du = \cos x \ \mathrm{d}x[/itex].

    The integral becomes: [itex]\displaystyle\int \dfrac {du}{u} \ \mathrm{d}x[/itex] and we integrate to get [itex]\ln |u|[/itex] (ignoring the constant of integration, for now -- reason: simplicity).

    We convert our result into a formula with respect to [itex]x[/itex]:
    [itex]\boxed{\ln |\sin x| + \mathcal{C}}[/itex]​

    My Question

    If we use Integration by Parts, however, with [itex]u=\dfrac{1}{\sin x}[/itex] and [itex]\mathrm{d}v = \cos x \ \mathrm{d}x[/itex] and
    [itex]\displaystyle\int u \ \mathrm{d}v = uv - \displaystyle\int v \ \mathrm{d}u,[/itex]​
    we get [itex]1=0[/itex]. I can show my solution, if needed. Why is this? (A friend of mine pointed this out to me)

    Thanks in advance! :)
     
  2. jcsd
  3. Nov 19, 2011 #2

    dynamicsolo

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    Homework Helper

    You get 1 = 0 only by rather slippery neglect of the arbitrary constant. The full integration-by-parts would have to read

    [tex]\int \frac{cos x dx}{sin x} = ( csc x )( sin x ) + \int (sin x) ( csc x cot x ) dx \mathbf{+ C } .[/tex]

    Someone had a post a month or so back (I don't know if I can find it again quickly) where the paradox there also turned on making a "1" appear out of nowhere. What you learn about indefinite integration is that finite numbers don't count for much...

    That sounds facetious, but you do find that, for example, certain trigonometric-powers integrals can give one form of an anti-derivative using one technique of integration, and a different (but equivalent) form of the anti-derivative using another method. When you apply trig identities to show the equivalence of the two results, you find that they differ by a small number. But since both results are general anti-derivatives with arbitrary constants, the small number is actually irrelevant (in a sense, it shows the difference in "vertical shift" of the two anti-derivative functions). So the specified constant value can be made to appear or disappear, depending on the choice of anti-derivative. One could say that arbitrary constants of integration "swallow" numerical constants.
     
    Last edited: Nov 19, 2011
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