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Take the integral of sqrt(64-x^2) dx.
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u=64-x^2?
Yes
Oh, no, this is not a u-sub problem. Didn't look at your problem that well.
I just knew it, then what's the strategy?
Bleh, trig sub. Which I'm not fond of, or good at. u = sqrt(64)sin(x), du/dx = sqrt(64)cos(x). Not too good at the next part...
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Replace x with sqrt(64)sin(x) I mean. The terms underneath the square root total 64cos^2(x)
Man, I'm losing it. It's something around those lines. Sorry for getting your hopes up lol.
use trigonometric substitution first take factor from the sqrt
\[\int\limits_{}^{}\sqrt{64-x^2}dx\] \[8\int\limits_{}^{}\sqrt{1-\frac{ x ^{2} }{ 64 }} dx\]
Thank you.
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\[8\int\limits_{}^{}\sqrt{1-(\frac{ x }{ 8 })^{2}} dx\]
np but can u complete it ?
@Idealist
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