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Homework Help: Definite integration

  1. Aug 25, 2012 #1
    1. The problem statement, all variables and given/known data
    [tex]\int_{0}^{4/ \pi} 3x^2 \cdot sin(\frac{1}{x})-x \cdot cos(\frac{1}{x})dx[/tex]

    2. Relevant equations

    3. The attempt at a solution
    I am stuck on this question because i really have no idea what should i be doing here with sin(1/x) and cos(1/x). I do not know how to integrate them. Even if i try to integrate by parts, i am stuck with these sin(1/x) and cos(1/x).

    Any help is appreciated.
  2. jcsd
  3. Aug 25, 2012 #2
    First, notice that you can split the integral into two different integrals. I will do one of them.


    The expansion of this integral involves the cosine integral, the case with sine is not very different. This makes me think that the integral has no elementary expansion, but I'm not sure. There might be a way to do it without splitting the integral.
  4. Aug 25, 2012 #3
    I have no idea what you meant by the "cosine integral".
    If i rewrite the expression as you said:
    [tex]\int_{\pi/4}^{\infty} \frac{3\sin(t)}{t^4}-\frac{\cos(t)}{t^3}\,dt[/tex]
    What should i be doing next?
  5. Aug 25, 2012 #4
    The indefinite integral has no representation via elementary functions. The definite integral has to be computed in some other way, e.g., via the residue theorem of complex analysis.
  6. Aug 25, 2012 #5
    Well, if you plug in the indefinite integral in wolfram alpha, the answer it shows is x3sin(1/x).
  7. Aug 25, 2012 #6
    I beg to differ ,
    This question can be tackled , by using integration by parts ,
    Let me give you a hint Pranav arora
    If you integrate 3sin(t)/t^4 , using integration by parts.
    Then one of two terms would be cos(t)/t^3 , which gets cancelled . and you are left with a simple integral which can be solved very easily :)
  8. Aug 25, 2012 #7
    Indeed. The addends cannot be integrated separately, but together the nicely cancel out the nasty part.
  9. Aug 25, 2012 #8
    Excellent, thanks kushan! That really worked! :smile:
  10. Aug 25, 2012 #9
    Your welcome :cool:
  11. Aug 25, 2012 #10


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    Homework Helper

    Nice trick, Kushan!

    Pranav, you should think of a trick if you are given such a problem.

    Try to consider the integrand as the derivative of a product (fg)=f'g+fg'.

    3x^2 is the derivative of x^3. cos() is the derivative of sin(), so find the derivative of sin(1/x)... It is -cos(1/x)/x^2. f'g+fg'=3x^2sin(1/x)-x^2cos(1/x)1/x^2= 3x^2sin(1/x)-xcos(1/x)

  12. Aug 25, 2012 #11
    Man, your trick made the question lot easier than before, thank you very much ehild! I will always try to use this wherever possible. :smile:

    As for the derivative of sin(1/x), it is -1/(x2)*cos(1/x).
    So according to the trick, the answer is x3sin(1/x). If i differentiate this, i get the same equation as mentioned in the question.
  13. Aug 25, 2012 #12


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    Homework Helper

    Right. Suck tricks are very useful in school, but hardly ever applicable in real life... Life is ugly.:tongue2:

  14. Aug 25, 2012 #13
    Yes this is really helpful , and these kind of question are very common here , they look impossible though at first look .
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