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Antiderivative Definite Integrals

  1. Feb 6, 2014 #1
    The problem statement, all variables and given/known data

    So I did an entire antiderivative, and ended with this part:

    sec(x)tan(x) + ln|sec(x) + tan(x)| + C

    I have to do this with the lower bound of -pi/3 and 0.
    When I do it, I should be getting 2√3 + ln(2+√3)

    But, I'm getting (0+0)-(2*-√3 + ln(2-√3))
    Which would give me 2√3-ln(2-√3)

    Will someone tell me where I'm going wrong?
    Does it have to do with the absolute value bars?

    Thanks in advance.
     
  2. jcsd
  3. Feb 6, 2014 #2
    Hi cathy!

    Both answers are correct.

    $$\ln(2-\sqrt{3})=\ln\left(\frac{1}{2+\sqrt{3}}\right)=-\ln(2+\sqrt{3})$$
     
  4. Feb 6, 2014 #3
    Oh! Silly error on my end.
    Thank you very much!
     
  5. Feb 6, 2014 #4
    Wait, maybe another silly error on my end, but I'm having a similar problem.

    The integral of secx dx with bounds of 0 and pi/4.
    The antideriv of sec x is ln(secx+tanx)
    so I plug everything in, and im getting ln(sqrt2-1)
    rather than the answer, which is ln(sqrt2+1)

    Am I doing something wrong?
     
  6. Feb 6, 2014 #5
    Yes. :)
    Not sure how you get this, what is the value of ##\tan(\pi/4)##?
     
  7. Feb 6, 2014 #6
    oh my, it is 1 and not 0.
    Sorry!

    Thank you so much, once again Pranav-Arora.
    I'm just a bit tired :)
    Thanks!
     
  8. Feb 6, 2014 #7
    Glad to help. :)
     
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