Question

ques

Answer 1

How may I solve the following integral?
\[\int\limits (16-x^2)^{\frac{3}{2}}dx\]

Answer 2

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

Answer 3

use trig substitute

Answer 4

x = 4 sint
dx = 4 cost dt
x^2 = 16 sin^2 t
16 -x^2 = 16 cos^2 t

Answer 5

(16 -x^2)^3/2 = 4^3 cos^3t ok?

Answer 6

no?? hehehe...

Answer 7

\[I=256 \int\limits \cos^4(t)dt\]

Answer 8

ok, go ahead

Answer 9

thanks a lot, let me try it now

Answer 10

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

Answer 11

and +C of course