# Help me with this simple integral.

1. Aug 9, 2012

### XtremePhysX

1. The problem statement, all variables and given/known data

Find:

2. Relevant equations

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

3. The attempt at a solution

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

2. Aug 9, 2012

### XtremePhysX

bump :)

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

Last edited: Aug 9, 2012
3. Aug 9, 2012

### LCKurtz

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. Aug 9, 2012

### tiny-tim

Hi XtremePhysX!

Try integrating by parts, or a trig substitution.

5. Aug 9, 2012

### XtremePhysX

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?

6. Aug 9, 2012

### tiny-tim

instead of using sin2θ, write it as 2sinθcosθ

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

7. Aug 9, 2012

### XtremePhysX

$$sin^{-1}\sqrt{x}-\frac{2sin(sin^{-1}\sqrt{x})cos(sin^{-1}\sqrt{x})}{2}=sin^{-1}\sqrt{x}-\frac{2(\sqrt{x})cos(sin^{-1}\sqrt{x})}{2}=sin^{-1}\sqrt{x}-\frac{2(\sqrt{x})cos(\sqrt{1-x})}{2}$$

Is this right?

8. Aug 9, 2012

### LCKurtz

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. Aug 10, 2012

### tiny-tim

(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) )

10. Aug 10, 2012

### Staff: Mentor

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.