Answer
$$\ln 2$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {x{{\sec }^2}xy}dA;\,\,\,\,\,\,\,R = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant \pi /3,\,\,\,\,0 \leqslant y \leqslant 1} \right\} \cr
& {\text{Convert to an iterated integral substituting the region }}R \cr
& = \int_0^{\pi /3} {\int_0^1 {x{{\sec }^2}xy} } dydx \cr
& = \int_0^{\pi /3} {\left[ {\int_0^1 {x{{\sec }^2}xy} dy} \right]} dx \cr
& {\text{solve the inner integral}}{\text{, treat }}x{\text{ as a constant}} \cr
& = \int_0^1 {x{{\sec }^2}xy} dy \cr
& {\text{integrating}} \cr
& = \left( {\tan xy} \right)_0^1 \cr
& {\text{evaluating the limits for the variable }}y \cr
& = \tan x\left( 1 \right) - \tan 0 \cr
& = \tan x \cr
& \int_0^{\pi /3} {\left[ {\int_0^1 {x{{\sec }^2}xy} dy} \right]} dx = \int_0^{\pi /3} {\tan x} dx \cr
& {\text{integrating}}{\text{, use }}\int {\tan u} du = - \ln \left| {\cos u} \right| + C \cr
& = \left( { - \ln \left| {\cos u} \right|} \right)_0^{\pi /3} \cr
& {\text{evaluate}} \cr
& = - \ln \left| {\cos \frac{\pi }{3}} \right| + \ln \left| {\cos 0} \right| \cr
& = - \ln \left( {\frac{1}{2}} \right) + \ln \left( 1 \right) \cr
& = \ln 2 \cr} $$
