1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration w/ trig sub

  1. Aug 18, 2013 #1
    1. The problem statement, all variables and given/known data

    integral of sqrt(x^2-36)/x)

    2. Relevant equations

    sqrt(x^2-a^2) = asec(u)
    Pythagorean identity

    3. The attempt at a solution
    I used trig sub on the x^2-36 and changed that to x=36sec(u) and dx= 36sec(u)tan(u). I simplified the square root in the numerator using sec^2(u)-1 = tan^2(u). This gave me (6tan(u)/36sec(u))(36sec(u)tan(u)) What trig identities do I need to proceed? Or did I use the wrong kind of trig sub? Any help is greatly appreciated in advance.
  2. jcsd
  3. Aug 18, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper

    hi jhahler! :smile:
    isn't that just tan2 ? :wink:
  4. Aug 18, 2013 #3


    User Avatar
    Homework Helper

    In your substitution, 'a' should be 6, not 36. Therefore your dx=6sec(u)tan(u).

    After your change your numbers, and simplify, you will need to use the same trig identity you used in the beginning with tan^2(u).
  5. Aug 19, 2013 #4


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    I don't understand why you do such a complicated substitution. Perhaps I don't understand your notation right. It's way easier to use
    [tex]x=6 \sin u.[/tex]
  6. Aug 19, 2013 #5


    User Avatar
    Science Advisor
    Homework Helper

    sec2 - 1 = tan2

    sin2 - 1 = minus cos2 :wink:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted