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ksaimouli

simplyfy

  • one year ago
  • one year ago

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  1. ksaimouli
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    how to get ln2 from that

    • one year ago
  2. ksaimouli
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    @zepdrix

    • one year ago
  3. ksaimouli
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    \[\int\limits_{\pi/6}^{\pi/2}\cot x dx\]

    • one year ago
  4. ksaimouli
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    i know that cotx =ln sinx +c

    • one year ago
  5. zepdrix
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    The integral of cotx? Ok good c:

    • one year ago
  6. ksaimouli
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    which will givw me ln1-ln(1/2)

    • one year ago
  7. ksaimouli
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    0-ln(1)-ln(2) gives me -ln2 but the book gives ln 2

    • one year ago
  8. zepdrix
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    \[\large \ln (a)-\ln(b)=\ln\left(\frac{a}{b}\right)\]

    • one year ago
  9. zepdrix
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    where is the 0- coming from? :o

    • one year ago
  10. ksaimouli
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    ln(1)=0

    • one year ago
  11. zepdrix
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    Oh you were doing something fancy, hmm

    • one year ago
  12. ksaimouli
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    |dw:1361499527104:dw|

    • one year ago
  13. zepdrix
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    Ok i see what you're doing. You just missed a negative in the middle.\[\large 0-\ln(1/2) \qquad = \qquad 0-\left(\ln1-\ln2\right)\]See where you missed it?

    • one year ago
  14. ksaimouli
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    ohhhhhhh godddddddd

    • one year ago
  15. zepdrix
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    heh

    • one year ago
  16. ksaimouli
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    thx buddy

    • one year ago
  17. ksaimouli
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    |dw:1361500220284:dw|

    • one year ago
  18. ksaimouli
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    how to do this @zepdrix

    • one year ago
  19. zepdrix
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    Write it like this, you might be able to see your `U substitution` easier.\[\large \int\limits (\ln x)^3 \left(\frac{1}{x}dx\right)\]

    • one year ago
  20. zepdrix
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    Hmm no that wouldn't do much for us.

    • one year ago
  21. ksaimouli
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    lnx

    • one year ago
  22. zepdrix
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    Think about the derivative of natural log and also the derivative of 1/x. Maybe one of those pieces could be our u.

    • one year ago
  23. zepdrix
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    Does the derivative of either one show up in the integral?

    • one year ago
  24. ksaimouli
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    if i take derivative of lnx then i get 1/x hmmm

    • one year ago
  25. zepdrix
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    Hmmm interesting.

    • one year ago
  26. ksaimouli
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    no idea!

    • one year ago
  27. zepdrix
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    So if we let \(\large u=\ln x\), taking the derivative of \(\large u\) gives us, \(\large du=\dfrac{1}{x}dx\) right?

    • one year ago
  28. ksaimouli
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    YES

    • one year ago
  29. zepdrix
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    Why is this in the biology section? lol

    • one year ago
  30. ksaimouli
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    oopsss i thought this is in math when i loged in i did not notice that

    • one year ago
  31. zepdrix
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    oh silly c:

    • one year ago
  32. zepdrix
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    So that would be the correctly substitution, because we have a suitable \(u\) and \(du\) that we can plug in. Understand how to plug them in?

    • one year ago
  33. ksaimouli
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    ohk i got it

    • one year ago
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