Answer
$\dfrac{4 \pi}{3}+\dfrac{\sqrt 3}{2}$
Work Step by Step
Let us consider that two continuous functions $f(x)$ and $g(x)$ with $f(x)\geq g(x)$ on the interval $[a,b]$ . The area (A) of the region bounded by the graph of $f(x)$ and $g(x)$ on the interval $[a,b]$ can be calculated as: $Area(A)=\int_a^b [f(x)-g(x)] \ dx$
The given curves are $y=\dfrac{\sec^2 x}{4}$ and $y=4 \cos^2 x$.The intersection points of these curves satisfy the given equations are: $\dfrac{\sec^2 x}{4}=4 \cos^2 x \implies x=\dfrac{\pi}{3} \ \text{and}\ x=-\dfrac{\pi}{3}$ . These points will be the lower and upper limit of the integration.
Thus, the area of the region is:
$A=\int_a^b [f(x)-g(x)] \ dx= \int_{-\dfrac{\pi}{3}}^{\dfrac{\pi}{3}} [4\cos^2 x-\dfrac{\sec^2 x}{4}] \ dx\\=2 \int_{0}^{\frac{\pi}{3}} [4(\dfrac{1+\cos 2x}{2})-\dfrac{\sec^2 x}{4}] \ dx\\ =[3x-\dfrac{x^2}{2}-\dfrac{2^x}{\ln 2}]_{0}^1\\ = 4(x+\dfrac{\sin 2x}{2})_0^{\pi/3}-\dfrac{2}{4}(\tan x)_0^{\pi/3}\\=\dfrac{4 \pi}{3}+\dfrac{\sqrt 3}{2}$
