Chapter 6 - Applications of the Integral - 6.1 Area Between Two Curves - Exercises - Page 286: 11

$\pi-2$

Since $2\geq \sec^2 x$ on $ -\pi/4\leq x\leq \pi/4$ Then \begin{align*} A&= \int_{-\pi / 4}^{\pi / 4}\left(2-\sec ^{2} x\right) d x\\ &=\left.(2 x-\tan x)\right|_{-\pi / 4} ^{\pi / 4}\\ &=\left(\frac{\pi}{2}-1\right)-\left(-\frac{\pi}{2}+1\right)\\ &=\pi-2 \end{align*}