Answer
$$\frac{1}{2}(\sqrt{2}-1)$$
Work Step by Step
Given $$\int_{0}^{\pi / 8} \sec 2 \theta \tan 2 \theta d \theta $$
Let $ u=2 \theta\ \ \Rightarrow \ \ du =2d \theta $ , at $ \theta=0\to u=0$ , at $ \theta=\pi / 8\to u=\pi / 4$
Then
\begin{align*}
\int_{0}^{\pi / 8} \sec 2 \theta \tan 2 \theta d \theta&=\int_{0}^{\pi / 4} \sec u \tan u\left(\frac{1}{2} d u\right)\\
&=\frac{1}{2}[\sec u]_{0}^{\pi / 4}\\
&=\frac{1}{2}\left(\sec \frac{\pi}{4}-\sec 0\right)\\
&=\frac{1}{2}(\sqrt{2}-1)
\end{align*}
