Question

Help getting started with the indefinite integral

\[\int \frac{2 \, dx}{\sqrt{16 - x^2}} =\]

Answer 1

i think this is the derivative of \(\arcsin(x)\) or at least it is close

Answer 2

if you rewrite it as
\[\frac{2}{4\sqrt{1-\left(\frac{x}{2}\right)^2}}\] then it is exactly that, except for the constant

Answer 3

ahh, it's a trig identity... I think I can get it from here... I see a 1-cos^2 x = sin^2 buried in the denominator..

I can get it from there I think thank you!

Answer 4

yw