Need help with a simple integral involving u substitution

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Homework Help Overview

The discussion revolves around evaluating the integral \(\int \sqrt{\frac{x}{1-x}}dx\) using various methods, including u-substitution and trigonometric substitution.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore u-substitution with \(u=1-x\) but express uncertainty about its effectiveness. Some suggest trying trigonometric substitution with \(x=\sin^2\theta\). Others mention integration by parts as a potential method.

Discussion Status

Several participants have shared their attempts and methods, with some providing guidance on alternative approaches. There is an ongoing exploration of different substitution techniques, but no consensus has been reached on a definitive solution.

Contextual Notes

Some participants express concerns about forum etiquette, particularly regarding post bumps, and seek clarification on the forum rules.

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Homework Statement



Find:

Homework Equations



\int \sqrt{\frac{x}{1-x}}dx

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



\int \sqrt{\frac{x}{1-x}}dx

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.
 
Hi XtremePhysX! :smile:

Try integrating by parts, or a trig substitution. :wink:
 
I found it =)

I used x=sin^2theta

and the answer is sin^{-1}\sqrt{x}-\frac{sin2(sin^{-1}\sqrt{x})}{2}how do i simplify it now?
 
instead of using sin2θ, write it as 2sinθcosθ :smile:

(but when you've done all that, start again and try it with integration by parts :wink:)
 
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}

Is this right?
 
XtremePhysX said:
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}

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.
 
(just got up :zzz:)

in other words cos(sin-1√x) = √(1 - x)

(because if y = sin-1√x, then √x = siny so x = sin2y so 1 - x = cos2y, so cosy = √(1 - x) :wink:)
 
  • #10
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.
 

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