Answer
$$\frac{{2\pi }}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{4ds}}{{\sqrt {4 - {s^2}} }}} \cr
& = 4\int_0^1 {\frac{{ds}}{{\sqrt {{{\left( 2 \right)}^2} - {s^2}} }}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 2 \cr
& = 4\left( {{{\sin }^{ - 1}}\left( {\frac{s}{2}} \right)} \right)_0^1 \cr
& = 4\left( {{{\sin }^{ - 1}}\left( {\frac{1}{2}} \right) - {{\sin }^{ - 1}}\left( {\frac{0}{2}} \right)} \right) \cr
& {\text{simplifying, we get:}} \cr
& = 4\left( {\frac{\pi }{6} - 0} \right) \cr
& = \frac{{2\pi }}{3} \cr} $$
