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

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  • #1
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[SOLVED] Find integral of sqrt((1-x)/(1+x))

Homework Statement


[tex]\int\sqrt{\frac{1-x}{1+x}}dx[/tex]


Homework Equations





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!
 

Answers and Replies

  • #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.
 
  • #3
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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.
 
  • #4
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The denominator should be sqrt(1-x^2) instead of just 1-x^2, correct? My easy algebra mistake!
 
  • #5
nicksauce
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Right... you can still solve with a trig sub, and a regular sub.
 
  • #6
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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!
 
  • #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.
 
  • #8
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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!
 
  • #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:

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