Integrating \sqrt{x-x^2} using Trig Substitution

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Homework Help Overview

The discussion revolves around the integration of the expression \(\sqrt{x-x^2}\), which falls under the subject area of calculus, specifically integral calculus. Participants are exploring various methods to tackle the integration problem.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using trigonometric substitution and u-substitution, with one participant expressing difficulty in integrating the resulting expressions. Others suggest completing the square or reformulating the integral to simplify the problem. There are questions about the necessity of using absolute values in the context of the integral's limits.

Discussion Status

The discussion is ongoing, with various approaches being explored. Some participants have provided suggestions that may lead to productive directions, while others are reflecting on their previous attempts and considering potential errors in their reasoning.

Contextual Notes

There is mention of the integral being evaluated over a specific range, [0, 1], which influences the discussion about the use of absolute values. Participants are also navigating the constraints of homework rules that may limit the methods they can employ.

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Homework Statement


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 can't make any progress on either.
 
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autodidude said:
[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?
 
Last edited:
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.
 
@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.
 
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.
 
try factorizing out the x... then use a substitution sqrt x = something... simplifies things a lot!
 

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