Answer
$$\frac{1}{2} - \frac{\pi }{8}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /8} {{{\tan }^2}2x} dx \cr
& {\text{Use the trigonometric identity ta}}{{\text{n}}^2}\theta + 1 = {\sec ^2}\theta \cr
& \int_0^{\pi /8} {{{\tan }^2}2x} dx = \int_0^{\pi /8} {\left( {{{\sec }^2}2x - 1} \right)} dx \cr
& = \frac{1}{2}\int_0^{\pi /8} {{{\sec }^2}2x\left( 2 \right)} dx - \int_0^{\pi /8} {dx} \cr
& {\text{Integrating}} \cr
& = \frac{1}{2}\left( {\tan 2x} \right)_0^{\pi /8} - \left( x \right)_0^{\pi /8} \cr
& = \left( {\frac{1}{2}\tan 2x - x} \right)_0^{\pi /8} \cr
& = \left( {\frac{1}{2}\tan 2\left( {\frac{\pi }{8}} \right) - \frac{\pi }{8}} \right) - \left( {\frac{1}{2}\tan 2\left( 0 \right) - \left( 0 \right)} \right) \cr
& {\text{Simplifying}} \cr
& = \frac{1}{2}\tan \left( {\frac{\pi }{4}} \right) - \frac{\pi }{8} - 0 \cr
& = \frac{1}{2} - \frac{\pi }{8} \cr} $$
