Integrate sqrt(x/2-x)dx

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  • #1
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How do I integrate sqrt(x/2-x)dx?
 

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  • #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] ?
 
  • #3
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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.
 
  • #4
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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.
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]
 
  • #5
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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].
 
  • #6
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When you said trigonometric substitution, you mean a^2-x^2=1-sin^2(x)?
 
  • #7
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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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