Answer
$\frac{\pi }{{16}} + \frac{1}{4}\ln \left( {\frac{{\sqrt 2 }}{2}} \right)$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /8} {x{{\sec }^2}2x} dx \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = x,{\text{ }}du = dx \cr
& dv = {\sec ^2}2xdx,{\text{ }}v = \frac{1}{2}\tan 2x \cr
& {\text{Integration parts formula }} \cr
& \int {\underbrace x_u} \underbrace {{{\sec }^2}2xdx}_{dv} = \underbrace {\left( x \right)}_u\underbrace {\left( {\frac{1}{2}\tan 2x} \right)}_v - \int {\underbrace {\left( {\frac{1}{2}\tan 2x} \right)}_v} \underbrace {dx}_{du} \cr
& \int {x{{\sec }^2}2xdx} = \frac{1}{2}x\tan 2x - \frac{1}{2}\int {\tan 2x} dx \cr
& \int {x{{\sec }^2}2xdx} = \frac{1}{2}x\tan 2x + \frac{1}{4}\ln \left| {\cos 2x} \right| + C \cr
& {\text{Therefore}}{\text{,}} \cr
& \int_0^{\pi /8} {x{{\sec }^2}2x} dx = \left[ {\frac{1}{2}x\tan 2x + \frac{1}{4}\ln \left| {\cos 2x} \right|} \right]_0^{\pi /8} \cr
& = \left[ {\frac{1}{2}\left( {\frac{\pi }{8}} \right)\tan 2\left( {\frac{\pi }{8}} \right) + \frac{1}{4}\ln \left| {\cos 2\left( {\frac{\pi }{8}} \right)} \right|} \right] - \left[ {0 + \frac{1}{4}\ln \left| {\cos 2\left( 0 \right)} \right|} \right] \cr
& = \left[ {\frac{\pi }{{16}}\left( 1 \right) + \frac{1}{4}\ln \left| {\frac{{\sqrt 2 }}{2}} \right|} \right] - 0 \cr
& = \frac{\pi }{{16}} + \frac{1}{4}\ln \left( {\frac{{\sqrt 2 }}{2}} \right) \cr
& \approx 0.109706 \cr} $$
