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Is this improper?

  1. May 18, 2008 #1
    I had this integral on my final exam and I was wondering at what point in the interval is the improper integral discontinuous?

    [tex]\displaystyle\int^{\pi/2}_{0} \frac{cos(x)dx}{sin^{2}(x) - 3sin(x) -4}[/tex]

    I know this is solved by partial fraction decomposition, but I don't see how the integral is improper?
     
    Last edited: May 18, 2008
  2. jcsd
  3. May 18, 2008 #2

    Dick

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    It's improper if the denominator vanishes, isn't it? But I don't think it does, so I don't think it is improper. Why do you think it is?
     
  4. May 18, 2008 #3
    I don't! But that was what the final asked of me -- evaluate the improper integral!
     
  5. May 18, 2008 #4

    Dick

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    Hard to say why they said that then. If you got it right, I wouldn't worry about it.
     
  6. May 18, 2008 #5
    Factor the denominator.
     
    Last edited: May 18, 2008
  7. May 18, 2008 #6
    Why I couldn't solve it using partials?

    1/[(sinx-4)*(sinx+1)] = A/(sin x - 4) + B/(sin x + 1 )

    so,
    A(sinx+1) + B(sinx-4) = 1

    Am I doing some stupid math?
     
  8. May 18, 2008 #7

    Dick

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    The OP already said it was known it could be solved by partials. Want's to know why it's 'improper'.
     
  9. May 18, 2008 #8
    I just wanted solve it for my own fun ;)
    I couldn't but matlab did:

    >> syms x;
    >> f = cos(x)/((sin(x))^2-3*(sin(x))-4);
    >> int(f,x,0,pi/2)

    ans =

    -3/5*log(2)+1/5*log(3) = -0.0851937465(approx using google calc)

    It's weird that I got "-0.1962" using other software: "Graph"...

    P.S. I don't think it was already known though
     
  10. May 18, 2008 #9
    It's improper bc of pi/2.
     
  11. May 18, 2008 #10

    Dick

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    The OP said "I know this is solved by partial fraction decomposition". It's not weird that you got "-0.1962", because that's the correct answer. What is strange is that "google calc" evaluated -3/5*log(2)+1/5*log(3) as -0.0851937465. That's wrong.
     
  12. May 18, 2008 #11

    Dick

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    Why??? What's going on with this thread?? The denominator is (-6) at x=pi/2!!!
     
    Last edited: May 18, 2008
  13. May 18, 2008 #12

    Dick

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    Ok. It's not that weird. Why did you use log10?
     
  14. May 18, 2008 #13
    oo thnx!
    matlab uses ln for log .. so that was the problem
     
  15. May 18, 2008 #14
    Oh sorry, I read my factorization incorrectly. I have like sinx+1 but read it as sinx-1, LOL.
     
  16. May 18, 2008 #15

    Dick

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    Ok, now what's rocomath's problem?
     
  17. May 18, 2008 #16

    Dick

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    So is everybody happy then?
     
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