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Find integral of sqrt((1-x)/(1+x))

  1. Mar 22, 2008 #1
    [SOLVED] Find integral of sqrt((1-x)/(1+x))

    1. The problem statement, all variables and given/known data
    [tex]\int\sqrt{\frac{1-x}{1+x}}dx[/tex]


    2. Relevant equations



    3. The attempt at a solution

    I have started by multiplying by [tex]\sqrt{1-x}[/tex] in the numerator and denominator. Then I separated the two fractions to get
    [tex]\int\frac{1}{1-x^2}dx[/tex][tex] -\int\frac{x}{1-x^2}dx[/tex] I'm stuck here! Any help is greatly appreciated!
     
  2. jcsd
  3. Mar 22, 2008 #2

    nicksauce

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    Well the first one can be solved by partial fractions (among other methods) and the second one can be solved by a simple substitution.
     
  4. Mar 22, 2008 #3
    Please check your multiplication for the denominator. Although wrong, the integrals you have are also easily integrable, first term with trigonometric substitution and the second term with ordinary variable substitution.
     
  5. Mar 22, 2008 #4
    The denominator should be sqrt(1-x^2) instead of just 1-x^2, correct? My easy algebra mistake!
     
  6. Mar 22, 2008 #5

    nicksauce

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    Right... you can still solve with a trig sub, and a regular sub.
     
  7. Mar 22, 2008 #6
    Thanks to all, does this look correct?

    This is what I got:

    [tex]=sin^-^1x+2\sqrt{1-x^2}+c[/tex]

    Thanks to everyone!
     
  8. Mar 22, 2008 #7

    D H

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    You are almost there. It's always a good idea to double check your integration by computing the derivative your result. If you do so, you will see you have a slight mistake.
     
  9. Mar 22, 2008 #8
    I believe my 2 should have cancelled out with the [tex]\frac{1}{2}[/tex]

    I'm now getting the answer [tex]=sin^-^1x+\sqrt{1-x^2}+c[/tex]

    After following your help, I did get the integral that I started with after I split it into two fractions. Thanks for your help and for catching my mistake!
     
  10. Mar 22, 2008 #9

    D H

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    Ahh. Much better.

    EDIT:
    Now that you know what the answer is supposed to be, you should go back to your derivation to see where you dropped a factor of 1/2 (or added a factor of 2).
     
    Last edited: Mar 22, 2008
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