Answer
$$\pi $$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^2 {\frac{{dp}}{{\sqrt {4 - {p^2}} }}} \cr
& {\text{The integrand is not continuous at }}x = \pm 2,{\text{ then}} \cr
& \int_{ - 2}^2 {\frac{{dp}}{{\sqrt {4 - {p^2}} }}} = \mathop {\lim }\limits_{a \to - {2^ + }} \int_a^0 {\frac{{dp}}{{\sqrt {4 - {p^2}} }}} + \mathop {\lim }\limits_{b \to {2^ - }} \int_0^b {\frac{{dp}}{{\sqrt {4 - {p^2}} }}} \cr
& {\text{Integrating}} \cr
& = \mathop {\lim }\limits_{a \to - {2^ + }} \left[ {{{\sin }^{ - 1}}\left( {\frac{p}{2}} \right)} \right]_a^0 + \mathop {\lim }\limits_{b \to {2^ - }} \left[ {{{\sin }^{ - 1}}\left( {\frac{p}{2}} \right)} \right]_0^b \cr
& = \mathop {\lim }\limits_{a \to - {2^ + }} \left[ {{{\sin }^{ - 1}}\left( 0 \right) - {{\sin }^{ - 1}}\left( {\frac{a}{2}} \right)} \right] + \mathop {\lim }\limits_{b \to {2^ - }} \left[ {{{\sin }^{ - 1}}\left( {\frac{b}{2}} \right) - {{\sin }^{ - 1}}\left( {\frac{0}{2}} \right)} \right] \cr
& = - \mathop {\lim }\limits_{a \to - {2^ + }} \left[ {{{\sin }^{ - 1}}\left( {\frac{a}{2}} \right)} \right] + \mathop {\lim }\limits_{b \to {2^ - }} \left[ {{{\sin }^{ - 1}}\left( {\frac{b}{2}} \right)} \right] \cr
& {\text{Evaluating the limit}} \cr
& = - \left[ {{{\sin }^{ - 1}}\left( {\frac{{ - 2}}{2}} \right)} \right] + \left[ {{{\sin }^{ - 1}}\left( {\frac{2}{2}} \right)} \right] \cr
& = \frac{\pi }{2} + \frac{\pi }{2} \cr
& = \pi \cr} $$
