## anonymous one year ago ques

1. anonymous

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

2. anonymous

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

3. Loser66

use trig substitute

4. Loser66

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

5. Loser66

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

6. Loser66

no?? hehehe...

7. anonymous

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

8. Loser66

9. anonymous

thanks a lot, let me try it now

10. anonymous

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

11. anonymous

and +C of course