Solving the Integral of (4-x)√(4x-x²)

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[tex]\int (4-x)\sqrt{4x-x^2}dx[/tex]

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

Steve
 
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Hmm... integration by parts, perhaps?
 
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It would be a lot easier if the integral were

[tex]\int {\left( {4 - 2x} \right)\sqrt {4x - x^2 } dx}[/tex].

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

Think about

[tex]\sqrt {4x - x^2 } = \sqrt {4 - \left( {x - 2} \right)^2 } = 2\sqrt {1 - \left( {\frac{{x - 2}}<br /> {2}} \right)^2 }[/tex]
 
Thanks TD, got it worked out now.

Steve