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Homework Help: 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
    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


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    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 :
    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
    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
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