Answer
$$ \frac{8}{3}$$
Work Step by Step
Given
$$ \int_{-\pi/4}^{\pi/4} \sec^4\theta d\theta= \int_{-\pi/4}^{\pi/4} (1+\tan^2 \theta)\sec^2\theta d\theta$$
Let
$$ u=\tan \theta \ \ \ \Rightarrow \ \ \ du = \sec^2 \theta d\theta$$
At $$ \theta= -\pi/4 \to u= -1, \ \ \theta= \pi/4\to u= 1$$
Then
\begin{aligned}
\int_{-\pi/4}^{\pi/4} \sec^4\theta d\theta&= \int_{-\pi/4}^{\pi/4} (1+\tan^2 \theta)\sec^2\theta d\theta\\
&= \int_{-1}^{1} (1+u^2)du\\
&= u +\frac{1}{3}u^3 \bigg|_{-1}^{1}\\
&= \frac{8}{3}
\end{aligned}
