# Another integral

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

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

Steve

## Answers and Replies

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James R
Homework Helper
Gold Member
Hmm... integration by parts, perhaps?

Last edited:
TD
Homework Helper
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

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

Thanks TD, got it worked out now.

Steve

TD
Homework Helper
Good