Answer
$$\frac{{{\pi ^2}}}{9} - 1$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /3} {\left( {2x - \sec x\tan x} \right)} dx \cr
& {\text{use power rule integration formulas from table 4}}{\text{.2}}{\text{.1}} \cr
& \int_0^{\pi /3} {\left( {2x - \sec x\tan x} \right)} dx = \left( {{x^2} - \sec x} \right)_0^{\pi /3} \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& \left( {{x^2} - \sec x} \right)_0^{\pi /3} = \left( {{{\left( {\frac{\pi }{3}} \right)}^2} - \sec \left( {\frac{\pi }{3}} \right)} \right) - \left( {{{\left( 0 \right)}^2} - \sec \left( 0 \right)} \right) \cr
& \sec \left( {\frac{\pi }{3}} \right) = 2{\text{ and sec}}\left( 0 \right) = 0 \cr
& = \frac{{{\pi ^2}}}{9} - 2 - 0 + 1 \cr
& = \frac{{{\pi ^2}}}{9} - 1 \cr} $$
