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Integrate sqrt(x/2-x)dx

  1. Aug 12, 2013 #1
    How do I integrate sqrt(x/2-x)dx?
     
  2. jcsd
  3. Aug 12, 2013 #2
    Is the problem:

    [itex]\int\sqrt{\frac{x}{2}-x}\space dx[/itex] or [itex]\int\sqrt \frac{x}{2-x}\space dx[/itex] ?
     
  4. Aug 12, 2013 #3
    I'm assuming it is the later. [itex] \int \sqrt{\frac{x}{2-x}} dx= \int \frac{\sqrt{x}}{\sqrt{2-x}} dx[/itex]

    Use the substitution [itex] u= \sqrt{2-x} [/itex] to turn the integral into [itex] -2 \int \sqrt{2-u^2} du [/itex], which you can do by trig substitution.
     
  5. Aug 12, 2013 #4
    Yes,it this one.I know U substitution, but I think I miss out something.How did you change [itex] u= \sqrt{2-x} [/itex] to turn the integral into [itex] -2 \int \sqrt{2-u^2} du [/itex]
     
  6. Aug 12, 2013 #5
    If [itex] u=\sqrt{2-x} [/itex], then [itex] \sqrt{x}=\sqrt{2-u^2} [/itex] and [itex] du=-\frac{dx}{2\sqrt{2-x}} [/itex].
     
  7. Aug 14, 2013 #6
    When you said trigonometric substitution, you mean a^2-x^2=1-sin^2(x)?
     
  8. Aug 14, 2013 #7
    More like the [itex] a^2-x^2=a^2(1-sin^2{\theta}) [/itex] that you will get when you set [itex] u^2=2sin^2\theta [/itex]
     
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