1. **State the problem:** We need to find the area of the region enclosed by the curves $y=\sec^2 x$ and $y=8\cos x$ over the interval $-\frac{\pi}{3} \le x \le \frac{\pi}{3}$. 2. **Understand the curves and interval:** - $y=\sec^2 x = \frac{1}{\cos^2 x}$ is always positive and increases as $x$ approaches $\pm \frac{\pi}{2}$. - $y=8\cos x$ oscillates between $8$ and $-8$. - The interval is symmetric about zero. 3. **Find points of intersection:** Set $\sec^2 x = 8 \cos x$: $$\frac{1}{\cos^2 x} = 8 \cos x \implies 1 = 8 \cos^3 x \implies \cos^3 x = \frac{1}{8} \implies \cos x = \frac{1}{2}.$$ 4. **Solve for $x$:** $$\cos x = \frac{1}{2} \implies x = \pm \frac{\pi}{3}$$ These are the endpoints of the interval, so the curves intersect at the boundaries. 5. **Determine which curve is on top between $-\frac{\pi}{3}$ and $\frac{\pi}{3}$:** At $x=0$: $$y=\sec^2 0 = 1, \quad y=8 \cos 0 = 8,$$ so $8 \cos x$ is above $\sec^2 x$ in the interval. 6. **Set up the integral for the area:** The area $A$ is $$A = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \left(8 \cos x - \sec^2 x\right) dx.$$ 7. **Calculate the integral:** $$\int 8 \cos x \, dx = 8 \sin x,$$ $$\int \sec^2 x \, dx = \tan x.$$ 8. **Evaluate definite integral:** $$A = \left[8 \sin x - \tan x \right]_{-\frac{\pi}{3}}^{\frac{\pi}{3}} = \left(8 \sin \frac{\pi}{3} - \tan \frac{\pi}{3}\right) - \left(8 \sin \left(-\frac{\pi}{3}\right) - \tan \left(-\frac{\pi}{3}\right)\right).$$ 9. **Use exact values:** $$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}, \quad \tan \frac{\pi}{3} = \sqrt{3}.$$ 10. **Substitute:** $$A = \left(8 \cdot \frac{\sqrt{3}}{2} - \sqrt{3}\right) - \left(8 \cdot \left(-\frac{\sqrt{3}}{2}\right) - (-\sqrt{3})\right) = \left(4 \sqrt{3} - \sqrt{3}\right) - \left(-4 \sqrt{3} + \sqrt{3}\right).$$ 11. **Simplify:** $$= (3 \sqrt{3}) - (-3 \sqrt{3}) = 3 \sqrt{3} + 3 \sqrt{3} = 6 \sqrt{3}.$$