# Another integral

1. Sep 25, 2005

### Stevecgz

$$\int (4-x)\sqrt{4x-x^2}dx$$

I'm uncertain how to find this integral, or even how to start it. Any guidence would be appreciated. Thanks.

Steve

2. Sep 25, 2005

### James R

Hmm... integration by parts, perhaps?

Last edited: Sep 25, 2005
3. Sep 26, 2005

### TD

It would be a lot easier if the integral were

$$\int {\left( {4 - 2x} \right)\sqrt {4x - x^2 } dx}$$.

If not, you'll need integration by parts indeed and I suspect you'll be getting an arcsin too

$$\sqrt {4x - x^2 } = \sqrt {4 - \left( {x - 2} \right)^2 } = 2\sqrt {1 - \left( {\frac{{x - 2}} {2}} \right)^2 }$$

4. Sep 27, 2005

### Stevecgz

Thanks TD, got it worked out now.

Steve

5. Sep 27, 2005

Good