1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    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!

Calc II substitution integration

  1. Sep 11, 2004 #1
    Hi all, I'm totaly lost on how to get started on this one.

    Integrate 1dx/(x^(1/2)(1-x)^(1/2))

    The solution manual jumps right through the substitution and ends up as

    Integrate (2u*du)/(u(1-u^2)^(1/2))

    I'm just not seeing it, little help please.

    Keith
     
  2. jcsd
  3. Sep 11, 2004 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you want to figure out what the solution manual did, try to identify the similarities between the two formulas, and identify what had to change to turn one into the other.


    The solution manual's way isn't the only way to do things anyways... what did you think would be a good thing to try to integrate this?
     
  4. Sep 11, 2004 #3
    Got it, thanks. My derivative taking was rustier then I thought.

    Let u=x^(1/2) which makes x=u^2

    here's where I kept screwing up
    du = 1/(2(x^(1/2)) dx
    so dx = 2 x^(1/2) or, as was declared above 2u. At that point the u's cancel and the 2 moves outside and it's in the form of a standard arcsin.

    You mentioned there might be another way to do it, I'm curious as to what that might be.
     
  5. Sep 11, 2004 #4
    algebra is a B1ach isn't it :-). I wish High school teachers took algebra more seriously because when students hit college, they are woefully unprepared for the amount of algebra they will need when learning higher level mathematics, and I can tell you that when I was in high school, I did not even thing about college level math and how much I would need to know from algebra.
     
  6. Sep 11, 2004 #5
    Another option is (though somewhat longer) to write the integrandum as 1/(x-x²)^(1/2)

    Then write as denominator sqrt(1/4 - (1/2-x)²), this equals sqrt (x - x²). Now just get the 1/4 out of the squareroot in order to come to : 1/2 sqrt(1-2²(1/2 -x)²) and now substitute u = 2(1/2-x) and you get a arcsin...

    Ok ??? Just for what it is worth

    regards
    marlon
     
  7. Sep 11, 2004 #6
    Thanks Marlon, I tried something like that at first, but it got messy so I bailed on it :). It did remind me of a few algebraic tricks I think I used to know about though.
     
  8. Sep 11, 2004 #7
    But you get what I am trying to say, kdinser???

    If so, then i am :biggrin:

    regards
    marlon
     
  9. Sep 16, 2004 #8

    Pythagorean

    User Avatar
    Gold Member

    word. I hated math in high school. I'm now wishing I could go back and spend less time on S, D, and R&R and more time on understanding complex relationships between numbers.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Calc II substitution integration
  1. Calc II Problem (Replies: 6)

  2. Calc integration (Replies: 8)

Loading...