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Partial fractions to determine antiderivative of sec x

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Derive a formula for the antiderivative of sec x using the identity that sec x= cos x/ (1-sin^2x). Use a substitution for sin x and then partial fractions. Then multiply the solution by (1+sin x)/ (1+sin x) to obtain the more familiar formula for the antiderivative.


    2. Relevant equations
    Known antiderivative of sec x: ln (tanx+secx)+c


    3. The attempt at a solution
    U= sinx, du =cosxdx
    intergral of du/1-u^2
    integral of A/(1-u) + integral of B/(1+u) = integral of 1/(1-u^2)
    A+B=1
    Au-Bu=0
    A= 1/2, B=1/2
    plugging back in and integrating I got 1/2(ln(1-sinx)+ln(1+sinx))... can this be simplified more?

    I am confused about how multiplying my solution by (1+sinx)/(1+sinx) is going to give me the formula I want... perhaps I made a math error? Please help!
     
  2. jcsd
  3. Nov 12, 2008 #2

    Dick

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    Yep, math error. The integral of 1/(1-u) isn't log(1-u). There's a sign error. Do you see it?
     
  4. Nov 12, 2008 #3
    I still don't see it...
     
  5. Nov 12, 2008 #4

    gabbagabbahey

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    As Dick said, you have a sign error....once you've corrected it you can combine your two log terms using the following property of logs :

    [tex]\ln (a) - \ln (b) = \ln \left( \frac{a}{b} \right)[/tex]
     
  6. Nov 12, 2008 #5

    gabbagabbahey

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    [tex]\frac{d}{du} \ln (1-u) = \frac{-1}{1-u}[/tex]

    :wink:
     
  7. Nov 12, 2008 #6
    Ok...so the integral in question should have been -ln(1-u)?
     
  8. Nov 12, 2008 #7

    Dick

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    The integral of 1/(1-u) is -log(1-u). Check it by differentiating. Use the chain rule.
     
  9. Nov 12, 2008 #8
    Oh, didn't see your last reply. Thanks. :)
     
  10. Nov 12, 2008 #9

    Dick

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    Sure.
     
  11. Nov 12, 2008 #10
    When I fixed my solution, I got 1/2*ln((1+sinx)/(1-sinx)).

    Does that look right?
     
  12. Nov 12, 2008 #11

    Dick

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    Yes. Can you finish it? Use the hint.
     
  13. Nov 12, 2008 #12

    gabbagabbahey

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    Looks fine, now what do you get when you multiply the argument of you logarithm by (1+sinx)/(1+sinx)?
     
  14. Nov 12, 2008 #13

    Dick

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    Slow night out there, gabbagabba?
     
  15. Nov 12, 2008 #14

    gabbagabbahey

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    All too often :wink:

    *quietly butts out of thread*
     
  16. Nov 12, 2008 #15
    I know 1/2ln ((1+sinx)/(1-sinx)) is equivalent to saying ln ((1+sinx)/(1-sinx))^1/2...
     
  17. Nov 12, 2008 #16

    Dick

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    Sure, can you get rid of those 1/2's by making the argument of the log a perfect square?
     
  18. Nov 12, 2008 #17

    Dick

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    No problem, but is was kind of echoy. We are saying exactly the same thing. Know anything about Banach spaces? There's a Carl140 problem way down the list I haven't had much time to think about.
     
  19. Nov 12, 2008 #18
    By completing the square? I'm not quite sure how to do that in this case...
     
  20. Nov 12, 2008 #19

    Dick

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    I told you, use the hint. Like gabbagabba said, multiply (1+sin(x))/(1-sin(x)) by (1+sin(x))/(1+sin(x)). Don't expand the numerator. Do expand the denominator.
     
  21. Nov 12, 2008 #20
    Ok, so I got the denominator to be cos^2(x), which I know is correct. My numerator is (1+sinx)*(1+sinx)^(1/2). I know this has to equal sin(x). I just don't see how...

    Thank you for being so patient with me. :smile:
     
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