Is the integral actually zero?

We note that $f(x) = \sqrt{1-\tan^2 x}$ is even; and that $f(0) = 1$ and $f\left(\pm \frac \pi 4\right) = 0$ . Therefore, the area of the shaded region is given by:

$\begin{aligned} I & = \int_{-\frac \pi 4}^\frac \pi 4 \sqrt{1-\tan^2 x} \ dx & \small \color{#3D99F6} \text{Since the integral is even.} \\ & = 2 \int_0^\frac \pi 4 \sqrt{1-\tan^2 x} \ dx & \small \color{#3D99F6} \text{Let }u = \tan x \implies du = \sec^2 x \ dx \\ & = 2 \int_0^1 \frac {\sqrt{1-u^2}}{u^2+1} \ du & \small \color{#3D99F6} \text{Let }u = \sin \theta \implies du = \cos \theta \ d\theta \\ & = 2 \int_0^\frac \pi 2 \frac {\cos^2 \theta}{\sin^2 \theta +1} \ d \theta \\ & = 2 \int_0^\frac \pi 2 \frac 1{\tan^2 \theta + \sec^2 \theta} \ d \theta \\ & = 2 \int_0^\frac \pi 2 \frac 1{2\tan^2 \theta + 1} \ d \theta & \small \color{#3D99F6} \text{Let }t = \tan \theta \implies dt = \sec^2 \theta \ d\theta \\ & = 2 \int_0^\infty \frac 1{(2t^2 +1)(t^2+1)} \ dt \\ & = 2 \int_0^\infty \left( \frac 2{2t^2 +1} - \frac 1{t^2+1} \right) dt \\ & = 2\bigg[\sqrt 2 \tan^{-1} 2t - \tan^{-1} t \bigg]_0^\infty \\ & = \left(\sqrt 2 - 1\right)\pi \\ & = \pi \tan \frac \pi 8 \end{aligned}$

$\implies A = \boxed{8}$