Answer
$$\frac{3}{2}-\sec \left( 1 \right)$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\left( {x - \sec x\tan x} \right)dx} \cr
& {\text{find the antiderivative }} \cr
& {\text{use formula }}\int {\sec x} \tan xdx = \sec x + C \cr
& \int_0^1 {\left( {x - \sec x\tan x} \right)dx} = \left( {\frac{{{x^2}}}{2} - \sec x} \right)_0^1 \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& \left( {\frac{{{{\left( 1 \right)}^2}}}{2} - \sec \left( 1 \right)} \right) - \left( {\frac{{{{\left( 0 \right)}^2}}}{2} - \sec \left( 0 \right)} \right) \cr
& {\text{simplify}} \cr
& = \left( {\frac{1}{2} - \sec \left( 1 \right)} \right) - \left( {0 - 1} \right) \cr
& = \frac{1}{2} - \sec \left( 1 \right) + 1 \cr
& = \frac{3}{2}-\sec \left( 1 \right) \cr} $$
