Need help with a simple integral involving u substitution

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Homework Statement



Find:

Homework Equations



[tex]\int \sqrt{\frac{x}{1-x}}dx[/tex]

The Attempt at a Solution



I tried to use u substitution with u=1-x but it did work.
 
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bump :)im really sorry, I am new here. Please excuse my actions.
where can i find the rules to read them?
 
Last edited:
XtremePhysX said:

Homework Statement



Find:

Homework Equations



[tex]\int \sqrt{\frac{x}{1-x}}dx[/tex]

The Attempt at a Solution



I tried to use u substitution with u=1-x but it did work.

XtremePhysX said:
bump :)

The moderators will likely slap your wrist for bumping within an hour of posting if they see it. I might try something like ##x=\sin^2\theta## and see what happens.
 
I found it =)

I used x=sin^2theta

and the answer is [tex]sin^{-1}\sqrt{x}-\frac{sin2(sin^{-1}\sqrt{x})}{2}[/tex]how do i simplify it now?
 
[tex]sin^{-1}\sqrt{x}-\frac{2sin(sin^{-1}\sqrt{x})cos(sin^{-1}\sqrt{x})}{2}=sin^{-1}\sqrt{x}-\frac{2(\sqrt{x})cos(sin^{-1}\sqrt{x})}{2}=sin^{-1}\sqrt{x}-\frac{2(\sqrt{x})cos(\sqrt{1-x})}{2}[/tex]

Is this right?
 
XtremePhysX said:
[tex]sin^{-1}\sqrt{x}-\frac{2sin(sin^{-1}\sqrt{x})cos(sin^{-1}\sqrt{x})}{2}=sin^{-1}\sqrt{x}-\frac{2(\sqrt{x})cos(sin^{-1}\sqrt{x})}{2}=sin^{-1}\sqrt{x}-\frac{2(\sqrt{x})cos(\sqrt{1-x})}{2}[/tex]

Is this right?

No. Call ##\theta = \arcsin({\sqrt x})##. You have ##\theta - \sin\theta \cos\theta## which is equal to ##\theta - \sqrt x \sqrt{1-\sin^2\theta}=\arcsin\sqrt x-\sqrt x \sqrt{1-x}##, which you can verify is correct by differentiating it.
 
XtremePhysX said:
bump :)


im really sorry, I am new here. Please excuse my actions.
where can i find the rules to read them?

I see you edited your post after the premature bump. Yes, please do not bump your post after just an hour -- the PF rules specify that you must wait at least 24 hours before making a single bump post.

EDIT -- And the Rules link is at the top of every PF page.