# Integrating with U-substitution

1. Sep 14, 2009

### zooboodoo

1. The problem statement, all variables and given/known data
Find the integral of (1-x)((2x-x^2)^.5) dx

2. Relevant equations

3. The attempt at a solution
I am kind of hazy on U substitutions but I thought that was the right way to go here:
let u=2x-x^2
du=2-2x
(1/2) Integral of (u)^.5 du < i wasn't sure if I went wrong here, Since i had (1-x) in the original equation and my du=2(1-x) can i put du in my substitution as long as I put (1/2) to balance the equation?

using this I got (2x-x^2)^(3/2)/(6) + C which was marked wrong. any insight would be appreciated

2. Sep 14, 2009

### njama

$$\int{\frac{1-x}{\sqrt{2x-x^2}}dx$$

You did the substitution well.

$$u=2x-x^2$$

$$du=(2-2x)dx=2(1-x)dx$$

So $dx=du/2(1-x)$

Now just substitute back in the integral (for dx and u)

$$\int{\frac{1-x}{\sqrt{u}}*\frac{du}{2(1-x)}}$$

3. Sep 14, 2009

### zooboodoo

oh no the initial integral is (1-x)Squareroot(2x-x^2) all in the numerator

4. Sep 14, 2009

### zooboodoo

$$\int{{1-x}{sqrt{2x-x^2}}dx$$

5. Sep 14, 2009

### tiny-tim

Hi zooboodoo!

(have a square-root: √ and try using the X2 tag just above the Reply box )
Yes, you've done everything perfectly, except you got the factor, 6, wrong at the end …

differentiate what you have (using the chain rule), and you'll see.

6. Sep 14, 2009

### zooboodoo

wah it was supposed to be /3 wasn't it :-x so frustraating

7. Sep 14, 2009

### tiny-tim

yeah … it's so easy to make a mistake like that!

always check any integration problem by differentiating your answer, and making sure you get back to where you started!