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

  • Thread starter Thread starter autodidude
  • Start date Start date
  • Tags Tags
    Integrate
autodidude
Messages
332
Reaction score
0

Homework Statement


Integrate \sqrt{x-x^2}

The attempt

I did a trig substitution, letting cos(\theta)=\frac{x}{sqrt(x)} and after some manipulation ended up with -2\int \ |sin(\theta)cos(\theta)|sin(\theta)cos(\theta) d\theta 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 2\int \ u\sqrt{u^2-u^4}du which I can't make any progress on either.
 
Physics news on Phys.org
autodidude said:
-2\int \ |sin(\theta)cos(\theta)|sin(\theta)cos(\theta) d\theta
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:

\int\sqrt{\frac{1}{4}-(x-}\frac{1}{2})^{2}dx

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

and see what you can get.
 
try factorizing out the x... then use a substitution sqrt x = something... simplifies things alot!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top