Answer
$$\frac{\pi }{2}$$
Work Step by Step
$$\eqalign{
& \int_0^3 {\frac{{dx}}{{\sqrt {9 - {x^2}} }}} \cr
& {\text{The integrand has an infinite discontinuity at }}x = 3. \cr
& {\text{Then by the definition of improper integrals: }} \cr
& \int_0^3 {\frac{{dx}}{{\sqrt {9 - {x^2}} }}} = \mathop {\lim }\limits_{b \to {3^ - }} \int_0^b {\frac{1}{{\sqrt {9 - {x^2}} }}} dx \cr
& \cr
& {\text{Integrating }}\int {\frac{1}{{\sqrt {9 - {x^2}} }}} dx,{\text{ using }}\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }}} = {\sin ^{ - 1}}\left( {\frac{x}{a}} \right) + C.{\text{ Let }}a = 3,{\text{ then}} \cr
& \mathop {\lim }\limits_{b \to {3^ - }} \int_0^b {\frac{1}{{\sqrt {9 - {x^2}} }}} dx = \mathop {\lim }\limits_{b \to {3^ - }} \left( {{{\sin }^{ - 1}}\left( {\frac{x}{3}} \right)} \right)_0^b \cr
& = \mathop {\lim }\limits_{b \to {3^ - }} \left( {{{\sin }^{ - 1}}\left( {\frac{b}{3}} \right) - {{\sin }^{ - 1}}\left( {\frac{0}{3}} \right)} \right) \cr
& = \mathop {\lim }\limits_{b \to {3^ - }} \left( {{{\sin }^{ - 1}}\left( {\frac{b}{3}} \right)} \right) \cr
& {\text{Evaluate the limit when }}b \to {3^ - } \cr
& = {\sin ^{ - 1}}\left( {\frac{3}{3}} \right) \cr
& = \frac{\pi }{2} \cr
& \cr
& {\text{Then}}{\text{,}} \cr
& \int_0^3 {\frac{{dx}}{{\sqrt {9 - {x^2}} }}} = \frac{\pi }{2} \cr} $$
