Answer
$\frac{\pi}{6}$
Work Step by Step
$\int_{0}^{\frac{1}{6}}\frac{3}{\sqrt (1-9x^{2})}dx$
The denominator $\sqrt (1-9x^{2})$ looks like $arcsin$. The formula to solve to $arcsin$ is:
$\int\frac{du}{\sqrt (a^{2}-u^{2})}=arcsin(\frac{u}{a})+C$
$u=3x$
$du=3dx$
$a=1$
$[arcsin(\frac{3x}{1})]_{0}^{\frac{1}{6}}$
$[arcsin(3(\frac{1}{6}))-arcsin(3(0))]$
$[arcsin(\frac{1}{2})-arcsin(0)]$
$[\frac{\pi}{6}-0]$
$\frac{\pi}{6}$
