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Integrate sqrt(x-x^2)

by autodidude
Tags: integrate, sqrtxx2
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autodidude
#1
Dec29-12, 03:58 PM
P: 333
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.
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haruspex
#2
Dec29-12, 04:19 PM
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Quote Quote by autodidude View Post
[tex]-2\int \ |sin(\theta)cos(\theta)|sin(\theta)cos(\theta) d\theta[/tex]
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?
Dick
#3
Dec29-12, 04:51 PM
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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.

autodidude
#4
Dec29-12, 05:18 PM
P: 333
Integrate sqrt(x-x^2)

@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.
mtayab1994
#5
Dec29-12, 07:06 PM
P: 584
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.
unscientific
#6
Dec30-12, 06:16 AM
P: 1,107
try factorizing out the x... then use a substitution sqrt x = something... simplifies things alot!


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