Answer
$s = 2\pi a$
Work Step by Step
$$\eqalign{
& r = 2a\cos \theta ,{\text{ }} - \frac{\pi }{2} \leqslant \theta \leqslant \frac{\pi }{2} \cr
& {\text{Find the Arc Length of a Polar curve}}{\text{, use the Theorem 10}}{\text{.14}} \cr
& s = \int_\alpha ^\beta {\sqrt {{{\left[ {f\left( \theta \right)} \right]}^2} + {{\left[ {f'\left( \theta \right)} \right]}^2}} } d\theta \cr
& {\text{Let }}r = f\left( \theta \right) = 4\sin \theta ,{\text{ }}\underbrace { - \frac{\pi }{2} \leqslant \theta \leqslant \frac{\pi }{2}}_{\alpha \leqslant \theta \leqslant \beta } \cr
& f\left( \theta \right) = 2a\cos \theta \cr
& f'\left( \theta \right) = - 2a\sin \theta \cr
& {\text{Therefore}}{\text{,}} \cr
& s = \int_{ - \pi /2}^{\pi /2} {\sqrt {{{\left[ {2a\cos \theta } \right]}^2} + {{\left[ { - 2a\sin \theta } \right]}^2}} } d\theta \cr
& s = \int_{ - \pi /2}^{\pi /2} {\sqrt {4{a^2}{{\cos }^2}\theta + 4{a^2}{{\sin }^2}\theta } } d\theta \cr
& s = \int_{ - \pi /2}^{\pi /2} {\sqrt {4{a^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)} } d\theta \cr
& s = \int_{ - \pi /2}^{\pi /2} {2a} d\theta \cr
& s = 2a\int_{ - \pi /2}^{\pi /2} {d\theta } \cr
& {\text{Integrate}} \cr
& s = 2a\left[ \theta \right]_{ - \pi /2}^{\pi /2} \cr
& s = 2a\left[ {\frac{\pi }{2} - \left( { - \frac{\pi }{2}} \right)} \right] \cr
& s = 2a\left( \pi \right) \cr
& s = 2\pi a \cr} $$
