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How to integrate?

  1. Jul 13, 2011 #1
    1. The problem statement, all variables and given/known data

    How do you integrate: 1/3(x^2(50 - x^2)^(3/2))

    2. Relevant equations



    3. The attempt at a solution

    I tried by parts with no luck :(
     
  2. jcsd
  3. Jul 13, 2011 #2
    Start out by taking out x^2 to get
    [tex]\frac{1}{x^2x^3(50x^{-2}-1)^{3/2}}[/tex]
     
  4. Jul 13, 2011 #3
    Sorry, it's kinda hard to read, I still don't follow.
     
  5. Jul 13, 2011 #4
    Take out (x^2) from the denominator's root.
    [tex]\int \frac{1}{3x^2(50-x^2)^{3/2}}dx = \int \frac{1}{3x^2x^3(50x^{-2}-1)^{3/2}} dx[/tex]
     
  6. Jul 13, 2011 #5

    SammyS

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    Do you mean: [itex]\displaystyle \int{\frac{1}{3x^2(50-x^{2})^{3/2}}}\,dx\ ?[/itex]

    What did you try for parts?

    A trig. substitution may work better.
     
  7. Jul 13, 2011 #6
    Oh, I see :) Very Nice, Thanks :)
     
  8. Jul 13, 2011 #7
    For parts I tried u=x, dv=x(50-x^2)^(3/2)

    How would you do it with trig sub?
     
  9. Jul 13, 2011 #8
    actually I'm stuck after factoring out x^2 because I get: 1/3 * (50/(x^2) - 1)^(3/2), but I don't know how to proceed.
     
  10. Jul 14, 2011 #9
    The x's don't cancel outside of the 3/2 root.
    The second step would be to do a substitution of everything inside the 3/2 root.
    [tex]\int \frac{1}{3x^2x^3(50x^{-2}-1)^{3/2}} dx[/tex]
    u = 50x^(-2) - 1
     
  11. Jul 14, 2011 #10
    The original equation is: (1/3)(x^2(50 - x^2)^(3/2)) when I take out x^2 I get: (1/3)(x^2*x^3(50/x^2 -1)^(1/3)) and if u= 50/x^2 -1 then du = -100/x^3 which doesn't help

    P.S. Latex code that you are using doesn't work and it makes it real hard to understand what you mean, if you could just use the keyboard I would get a better idea of what you mean.
     
  12. Jul 14, 2011 #11
    Ah, sorry I thought it was all in the denominator.
    For that one all I think of is a trig substitution, which isn't my best area.

    But looking at it I'd think you want to try x = 5 sqrt(2) sin(u) so that you can get rid of the root into a cosine.
     
  13. Jul 14, 2011 #12
    Thanks I'll give it a shot.
     
  14. Jul 14, 2011 #13
    Yeah, it's a pretty nasty problem, I don't know why they would include it in our book without explanation.
     
  15. Jul 14, 2011 #14

    SammyS

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    So, it really is: [itex]\displaystyle \int{\frac{1}{3}x^2(50-x^{2})^{3/2}}\,dx\ ?[/itex]

    For parts, I would use u = (1/3)(50-x^{2})^{3/2} & dv = x2dx.
     
  16. Jul 14, 2011 #15
    Ok I'll try that as well, thanks.
     
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