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Integrate sqrt(x-x^2)

  1. Dec 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Integrate [tex]\sqrt{x-x^2}[/tex]

    The attempt

    I did a trig substitution, letting [tex]cos(\theta)=\frac{x}{sqrt(x)}[/tex] and after some manipulation ended up with [tex]-2\int \ |sin(\theta)cos(\theta)|sin(\theta)cos(\theta) d\theta[/tex] which I have no idea how to integrate.

    If I make a u-substitution and let u=cos(theta) rather than simplify to get the above, I get [tex]2\int \ u\sqrt{u^2-u^4}du[/tex] which I cant make any progress on either.
     
  2. jcsd
  3. Dec 29, 2012 #2

    haruspex

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    The original integral must be over a range in [0, 1]. This means you can restrict theta to [0, pi/2], allowing you to drop the modulus function, leaving sin2cos2. Can you solve it from there?
     
    Last edited: Dec 29, 2012
  4. Dec 29, 2012 #3

    Dick

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    The more common way to do a problem like this is to complete the square inside the radical then substitute. I think it goes a bit easier that way.
     
  5. Dec 29, 2012 #4
    @haruspex: Yeah, I tried that and when I got the incorrect answer, I went back and saw that I overlooked the fact that you need to insert the modulus wheen rooting a square. Will try again in case I made an error though.

    @Dick: Thanks, I'll see where I can get with that.
     
  6. Dec 29, 2012 #5
    Like Dick said. Look at it like this try to reformulate it so you get something like this:

    [tex]\int\sqrt{\frac{1}{4}-(x-}\frac{1}{2})^{2}dx[/tex]

    and substitute u : [tex]u=x-\frac{1}{2};dx=du[/tex]

    and see what you can get.
     
  7. Dec 30, 2012 #6
    try factorizing out the x... then use a substitution sqrt x = something... simplifies things alot!
     
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