$\displaystyle \int \frac{dx}{\sqrt{9-x^2}}$
I know this is related to Inverse Tringemetric functions, so the Integral should somehow end $\displaystyle {sin^{-1}}x$
$\displaystyle \int \frac{3{\cos u}du}{\sqrt{9-9 \sin^{2}u}}$
$\displaystyle \int \frac{1}{\sqrt{9-9 \sin^{2}u}}{3{\cos u}du}$
But please stay with me, this is in a chapter of Tringometric functions
and
$\displaystyle \frac{d}{dx} (\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}$
This should somehow relate, now I'm stuck with 9 that I want to become 1