Answer
$\displaystyle \frac{4\pi}{3}$
Work Step by Step
On the interval $[-\displaystyle \frac{\pi}{3},\frac{\pi}{3}]$, the graph of $y=\displaystyle \frac{1}{2}\sec^{2}t$ is above the graph of $y=-4\sin^{2}t$,
The area between the two graphs over [a,b], where one graph is above the other, is given with $\displaystyle \int_{a}^{b}(y_{above}-y_{below})dt$.
Thus,
$A=\displaystyle \int_{-\pi/3}^{\pi/3}[\frac{1}{2}\sec^{2}t-(-4\sin^{2}t)]dt=\int_{-\pi/3}^{\pi/3}[\frac{1}{2}\sec^{2}t+4\sin^{2}t]dt$
$=\displaystyle \frac{1}{2}\int_{-\pi/3}^{\pi/3}\sec^{2}tdt+4\int_{-\pi/3}^{\pi/3}\sin^{2}tdt$
The first integral is not a problem as $\displaystyle \frac{d}{dt}[\tan t]=\sec^{2}t$
In the second, we use the double angle formula for $\cos 2t$
$\cos 2t=\cos^{2}x-\sin^{2}t=-(1-\sin^{2}t)-\sin^{2}t=1-2\sin^{2}t$,
We transform $\displaystyle \sin^{2}t= \frac{1}{2}(1-\cos 2t)$
$=\displaystyle \frac{1}{2}[\tan t]_{-\pi/3}^{\pi/3}+2\int_{-\pi/3}^{\pi/3}dt-2\int_{-\pi/3}^{\pi/3}\cos 2tdt$
For the last integral : $\left[\begin{array}{lll}
u=2x, & & du=2dx\\
& & dx=du/2\\
\text{Borders:} & & \\
x=-\pi/3 & \rightarrow & u=-2\pi/3\\
x=\pi/3 & \rightarrow & u=2\pi /3
\end{array}\right]$
$=\displaystyle \frac{1}{2}[\sqrt{3}+\sqrt{3}]_{-\pi/3}^{\pi/3}+2[\frac{\pi}{3}-(-\frac{\pi}{3})]- 2\int_{-2\pi/3}^{2\pi/3}\cos u\frac{du}{2}$
$= \displaystyle \sqrt{3}+\frac{4\pi}{3}-1\cdot[\sin u]_{-2\pi/3}^{2\pi/3}$
$= \displaystyle \sqrt{3}+\frac{4\pi}{3}-[\frac{\sqrt{3}}{2}-(-\frac{\sqrt{3}}{2})]$
$= \displaystyle \sqrt{3}+\frac{4\pi}{3}-\sqrt{3}$
= $\displaystyle \frac{4\pi}{3}$
