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Evaluate the Indefinite Integral of sqrt(16x-x^2) dx
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Complete the square of the integrand to get 64-(x-8)^2
Then let u = x-8 and du = dx
Now you have; \[\int\limits_{?}^{?}\sqrt{64-u^2}du\]
Substitute u = 8sin(s) and du = 8 cos(s)ds
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