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How may I solve the following integral? \[\int\limits (16-x^2)^{\frac{3}{2}}dx\]
I know for \[\int\limits (a^2-x^2)^\frac{1}{2}dx=\frac{x}{2}.(a^2-x^2)^\frac{1}{2}+\frac{a^2}{2}\sin^{-1}(\frac{x}{a})\] But this is cubed now
use trig substitute

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Other answers:

x = 4 sint dx = 4 cost dt x^2 = 16 sin^2 t 16 -x^2 = 16 cos^2 t
(16 -x^2)^3/2 = 4^3 cos^3t ok?
no?? hehehe...
\[I=256 \int\limits \cos^4(t)dt\]
ok, go ahead
thanks a lot, let me try it now
Ok so I'm getting a really weird answer \[I=96 \sin^{-1}({\frac{x}{4}})+8x\sqrt{16-x^2}+\frac{x\sqrt{16-x^2}(8-x^2)}{32}\] Now I can't say for sure how much accurate this answer is, but more or less I now understand how to do that integral thanks @Loser66
and +C of course

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