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Homework Help: Integral of cos and ln

  1. Apr 8, 2007 #1
    Please, how can I solve this?

    ∫ cos x ln x dx

    I get this:

    ln x sin x - ∫sin x/x dx

    but how do I continue from here?

    Thanks in advance...
  2. jcsd
  3. Apr 8, 2007 #2
    The antiderivative of sin(x)/x isn't expressible in terms of elementary functions so perhaps it would be better to change the role of u and dv in your integration by parts.

    Edit: At least, I think that's the case. Couldn't hurt to try anyway.
  4. Apr 8, 2007 #3


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    That won't change anything -- one cannot be expressed in terms of elementary functions iff the other cannot be expressed as well.
  5. Apr 8, 2007 #4
    look at it as an equation, and you need to integrate by parts at least twice
  6. Apr 9, 2007 #5


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    No, this won't help. Even Wolfram gives an answer with Si(x) in it - the integral of Sinx/x.
  7. Apr 9, 2007 #6
    So, can't sinx/x be integrated?
  8. Apr 9, 2007 #7


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    Yes, of course it can- its integral is Si(x)! It cannot, however, be integrated in terms of elementary functions.
  9. Apr 9, 2007 #8
    It seems difficult to continue from ln x sin x - ∫sin x/x dx ...

    Thanks to averyone who posted. I'll tell you if something different appears.

    Thanks again.
  10. Apr 9, 2007 #9
    It is impossible to continue without introducing "special" functions or series expansions (from which you won't be able to obtain closed forms). So, play with it for a while, but don't spend too much time on it :smile:.
  11. Apr 10, 2007 #10

    Gib Z

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    If you really don't want Si(x), heres your only alternative:

    [tex]\frac{\sin x}{x} = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n+1)!}[/tex].

    Integrate that, and there you go.
  12. Apr 10, 2007 #11
    And don't forget this one:
    [tex] \frac{sin(x)}{x} = sinc(x)[/tex]
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