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Integral of Logarithms + Trig Functions

  1. Sep 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Sec(x)/((ln(tan(x)+sec(x))^1/2)

    We were instructed to find the integral

    2. Relevant equations

    Here is a link to the wolfram solution, i dont understand the steps they took


    http://www.wolframalpha.com/input/?i=Integral+of+Sec%28x%29%2F%28%28ln%28sec%28x%29%2Btan%28x%29%29^1%2F2%29%29




    3. The attempt at a solution

    I understand that using the substitution method using u = Ln(tan(x)+sec(x))

    du = Sec(x)^2+tan(x)Sec(x)/Sec(x)+Tan(x) dx

    I do not understand how to substitute this and get 1/((u)^1/2) which is according to wolfram, I dont understand how the substitution method eliminates the Sec from the numerator, Later they evaluate that 1/(u^(1/2)) as 2((u)^1/2) can someone please explain these intermediate steps?
     
  2. jcsd
  3. Sep 6, 2011 #2

    micromass

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    Observe that du=sec(x)dx. Can you do it now?
     
  4. Sep 6, 2011 #3
    So, du = Sec(x)^2+tan(x)Sec(x)/Sec(x)+Tan(x) dx, reduces to sec(x)dx?
     
  5. Sep 6, 2011 #4

    micromass

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    Yes, just factor sec(x) in the numerator.
     
  6. Sep 6, 2011 #5
    Ah! Thank you, ive been looking at it for so long i didnt even see that! the only thing left is that i dont understand how the integral of 1/(u^1/2) is 2(u^1/2)
     
  7. Sep 6, 2011 #6

    micromass

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    In, general, what is the integral of un?? You must have seen a special formula for that...
     
  8. Sep 6, 2011 #7
    HaHa...wow I see it all now, I need to sleep, Thank you so much!
     
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