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Sqrt Integration

  1. Apr 30, 2009 #1
    1. The problem statement, all variables and given/known data
    hi guys ,, how are you all ,,
    i got another problem -_-

    integral(1/sqrt(4 x-x^2), x)

    2. Relevant equations



    3. The attempt at a solution
    i have no idea , i tried and lift it up and make it (4x-x^2)^-0.5 and i even took x as common factor x^-0.5 * (4-x)^-0.5 but still couldn't do anything next ,, any ideas ??
     
  2. jcsd
  3. Apr 30, 2009 #2

    Tom Mattson

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    You need to complete the square under the radical sign.
     
  4. Apr 30, 2009 #3
    i don't think they taught us how to change it to radicals ,, can you tell me what's the name of the chapter that teach this thing ? or at least give me the name of the method so i can look up for it
     
  5. Apr 30, 2009 #4

    djeitnstine

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    Lord dark, complete the square. That will get you to the next step

    Edit: tom beat me to it

    edit2: dark, he did not say change it radicals, but to complete the square
     
  6. Apr 30, 2009 #5
    lol ,, got the idea ,, thanks guys ,, i'll try then i'll give you the results
     
  7. Apr 30, 2009 #6
    i tried to solve and i reached here ,, 1/sqrt(4x-x^2) = 1/abs(x)*sqrt(4-x^2) and i know it has something with inv(sec x) but the problem now is 4 ,, i don't know how to make an equation that differentiable and give 1/sqrt(4x-x^2) from inv (sec x),, so i read in wikipedia (http://upload.wikimedia.org/math/6/3/f/63ff35f597f9092362ae154641cfe48a.png)
    so i think i get this answer: 0.5*inv(sec (x/2))+c ,, is it right ??
     
  8. Apr 30, 2009 #7

    djeitnstine

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    No that is wrong

    When completing the square, you make it in the form [tex]\left( x \pm a\left)^2 - b [/tex]

    So you will be looking at [tex]\int{\frac{dx}{\sqrt{\left( x \pm a\left)^2 - b}}}[/tex] which becomes an easy trig substitution
     
    Last edited: Apr 30, 2009
  9. May 1, 2009 #8
    ok ,, i think i did it right this time : integral[1/sqrt(4-(x-2)^2)] and i know its like the inverse of sin(x) [http://upload.wikimedia.org/math/2/e/8/2e805e2888bd24e7f61e54a002462204.png] but , i have 4 in square root instead of 1 what should i do ?? i tried and applied the previous way i learn from Wikipedia [0.5*inv(sin((x-2)/2))] but when i check i get wrong answer
     
    Last edited: May 1, 2009
  10. May 1, 2009 #9

    HallsofIvy

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    [tex]\frac{1}{\sqrt{4- (x-2)^2}}= \frac{1}{\sqrt{4(1-\frac{(x-2)^2}{4}}}= \frac{1}{2}\frac{1}{\sqrt{1- \frac{(x-2)^2}{4}}}[/tex]
     
  11. May 1, 2009 #10
    Lol ,, i think am stupid now -_- ,, i got it until the second phase but i didn't think of getting 4 out of square root ,, thanks very much guys for the help
     
    Last edited: May 1, 2009
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