$\begin{aligned} & \int_0^{\frac{\pi}{4}} \sec ^4 x \mathrm{~d} x=\int_0^{\frac{\pi}{4}}\left(\tan ^2 x+1\right) \sec ^2 x \mathrm{~d} x \\ & =\left[\frac{\tan ^3 x}{3}+\tan x\right]_0^{\frac{\pi}{4}} \\ & =\frac{1}{3}+1 \\ & =\frac{4}{3}\end{aligned}$