Integrating with U-substitution

[zooboodoo]

Homework Statement

Find the integral of (1-x)((2x-x^2)^.5) dx

Homework Equations

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

 

[njama]

Your original integral is:

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

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)

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

 

[zooboodoo]

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

 

[zooboodoo]

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

 

[tiny-tim]

Hi zooboodoo!

(have a square-root: √ and try using the X² tag just above the Reply box )

… (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

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.

 

[zooboodoo]

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

 

[tiny-tim]

… so frustraating

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!