1. Limited time only! Sign up for a free 30min personal 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!

Tricky integral

  1. Nov 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Solve dW/dx=W(4-2W)^(1/2)



    2. Relevant equations



    3. The attempt at a solution
    dW/[W(4-2W)^(1/2)]=dx
    dW/[W(4-2W)^(1/2)]=x+C
    I'm unsure of whether to use a u-sub, partial fractions, or a trig substition to integrate. I looked in the back of the book and the answer has a sech in the answer but how to get there?
     
  2. jcsd
  3. Nov 6, 2009 #2
    First try u = 4 - 2W
     
  4. Nov 6, 2009 #3
    I did that and was left with dW/[Wu^(1/2)]. I know du would give me -2. I thought about using du/[(u-4)u^(1/2)], but wasn't sure what to do from there.
     
  5. Nov 6, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Then you didn't do it right! When you substitute, you cannot leave the "old" variable in the integral, not even in the differential. If u= 4-2W, then 2W= 4- u so W= (4-u)/2 and dW= (-1/2)du.
     
  6. Nov 6, 2009 #5
    You need to "replace" or substitute all W's with u's in the integrand. Using u = 4 - 2W, solve for W and substitute that for W in the integrand. You also need du in your integrand, not dW.
     
  7. Nov 6, 2009 #6
    So, I'm left with :
    -du/[(4-u)u^(1/2)]
    Then I'm stuck
     
  8. Nov 6, 2009 #7
    Use another substitution that gets rid of the square root.
     
  9. Nov 6, 2009 #8
    So, using a=u^(1/2) and a^2=u, then du=2ada and 4-u=4-a^2
    -2ada/[(4-a^2)a]=2da/(a^2-4). This leads to a trig answer because I think there is a trig law with a^2+c^2?
     
  10. Nov 6, 2009 #9
    You could use a trig substitution; however, partial fractions would be better.
     
  11. Nov 6, 2009 #10
    So, I got 1/2ln[(4-2w)^(1/2)-2]-1/2ln[(4-2w)^(1/2)+2]=x+c
     
  12. Nov 6, 2009 #11
    After checking the answer on wolframalpha, it looks like you're missing a couple numbers in the ln's.
     
  13. Nov 6, 2009 #12
    I went from:
    partial fractions A=1/2, B=-1/2
    1/2ln(a-2)-1/2ln(a+2)
    1/2ln(u^1/2-2)-1/2ln(u^1/2+2)
    So, I got 1/2ln[(4-2w)^(1/2)-2]-1/2ln[(4-2w)^(1/2)+2]=x+c
    Where would I have gone wrong to need these extra numbers?
     
  14. Nov 6, 2009 #13
    Differentiating your result does give the same thing you started with, so your answer is correct.
     
  15. Nov 7, 2009 #14
    ok thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Tricky integral
  1. Tricky integral (Replies: 3)

  2. Tricky integral (Replies: 7)

  3. Tricky integral (Replies: 3)

  4. Tricky integrals (Replies: 4)

  5. Tricky integral (Replies: 1)

Loading...