Answer
$\frac{8}{3}\sqrt 3 $
Work Step by Step
$$\eqalign{
& \text{Let }I=\int_{\pi /6}^{\pi /3} {4{{\sec }^2}y} dy \cr
& {\text{Pull out the constant}} \cr
& I = 4\int_{\pi /6}^{\pi /3} {{{\sec }^2}y} dy \cr
& {\text{Integrate}}{\text{, recall that }}\int {{{\sec }^2}x} dx = \tan x + C \cr
& I= 4\left[ {\tan y} \right]_{\pi /6}^{\pi /3} \cr
& {\text{Using the fundamental theorem of calculus}}{\text{, part 2}} \cr
& I = 4\left[ {\tan \left( {\frac{\pi }{3}} \right) - \tan \left( {\frac{\pi }{6}} \right)} \right] \cr
& {\text{Simplify}} \cr
& I= 4\left( {\sqrt 3 - \frac{{\sqrt 3 }}{3}} \right) \cr
& = \frac{8}{3}\sqrt 3 \cr} $$
