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\[\cos^2 \theta = 1 - \sin^2 \theta\]

ok

Do you see how that helps us here?

yes because that will get you \[\sqrt{64\cos ^{2}\theta }\]

and then dx=8cos t dt
t for theta

Good. Keep going!

now this is where i'm getting kind of lost

and then i'll work on integrating that

Sure, one sec

Actually I think I spaced out a bit.
I'm just getting \(\int d\theta\)

no the answer is something weird
\[\frac{ x }{ 2 }\sqrt{64-x ^{2}}+32\sin^{-1} (\frac{ x }{ 8 })+C\]

Ok, that looks like the result of integration by parts after all.
So how did you get that? :D

back of the book lol

that's why I'm having trouble I can't get there

correct

ok

And yes I forgot to write the other 8.. Errr...

Yeah I missed my 2.