Need help with a simple integral involving u substitution

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In summary, the conversation is about finding the integral of \sqrt{\frac{x}{1-x}} using various methods such as trigonometric substitution and integration by parts. The final solution is simplified to sin^{-1}\sqrt{x}-\sqrt{x}\sqrt{1-x}. The poster also asks about the rules of the forum.
  • #1
XtremePhysX
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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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  • #2
bump :)im really sorry, I am new here. Please excuse my actions.
where can i find the rules to read them?
 
Last edited:
  • #3
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.
 
  • #4
Hi XtremePhysX! :smile:

Try integrating by parts, or a trig substitution. :wink:
 
  • #5
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?
 
  • #6
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:)
 
  • #7
[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?
 
  • #8
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.
 
  • #9
(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.
 

1. How do you solve an integral?

Solving an integral involves finding the antiderivative of a given function and evaluating it at the upper and lower limits of integration. This process is also known as integration or antidifferentiation.

2. What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration and yields a single numerical value, while an indefinite integral does not have limits and results in a function.

3. How do you use substitution to solve an integral?

Substitution is a technique used to simplify an integral by replacing a variable with a new one. This new variable can then be integrated more easily, and the answer can be converted back to the original variable.

4. Can you use integration by parts to solve any integral?

No, integration by parts can only be used for certain types of integrals, such as those that involve products of functions. It is not a universal method for solving all integrals.

5. How do you know when to use different integration techniques?

The choice of integration technique depends on the form of the integrand and the given limits of integration. It is important to be familiar with a variety of integration methods and to choose the most appropriate one for each problem.

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