Answer
$\frac{\pi}{2}$
Work Step by Step
$\int{\frac{dx}{\sqrt {9-x^{2}}}}$ = $sin^{-1}x$
$\int_0^R{\frac{dx}{\sqrt {9-x^{2}}}}$ = $sin^{-1}(\frac{R}{3})$
$\int_0^3{\frac{dx}{\sqrt {9-x^{2}}}}$ = $\lim\limits_{R \to {3^{-}}}$$sin^{-1}(\frac{R}{3})$ = $\frac{\pi}{2}$
