Question

Integral from 0 to 4*sqrt(2) of 1/sqrt(16-x^2). I'll rewrite below.

Answer 1

\[\int\limits_{0}^{4\sqrt{2}}\frac{ dx }{ \sqrt{16-x^2} }\] I think I am trying to use the trig substitution of sin here.

Answer 2

\[x=4\sin (\theta)\]

Answer 3

let x = 4sinθ

Answer 4

Is it \[(16 - 4\sin(\theta))^{1/2}\]

Answer 5

Ahh wait. It's \[(16-4(\sin (\theta))^{2})^{1/2}\]

Answer 6

x = 4sinθ
dx = 4cosθ dθ

x = 4sinθ then
sqrt(16-16sin^2 θ)=sqrt(16(1-sin^2 θ))=sqrt(16cos^2 θ)= 4cosθ

Answer 7

see your integral becomes