## University Calculus: Early Transcendentals (3rd Edition)

The area of the shaded region is $4\pi/3$.
The shaded region is bounded above by the curve $y=\frac{1}{2}\sec^2t$ and below by the curve $y=-4\sin^2t$ and runs from $x=-\pi/3$ to $x=\pi/3$. Therefore, according to the definition, the area of the shaded region is $$A=\int^{\pi/3}_{-\pi/3}\Big(\frac{1}{2}\sec^2t-(-4\sin^2t)\Big)dt$$ $$A=\frac{1}{2}\int^{\pi/3}_{-\pi/3}\sec^2tdt+4\int^{\pi/3}_{-\pi/3}\sin^2tdt$$ $$A=\frac{1}{2}\int^{\pi/3}_{-\pi/3}\sec^2tdt+4\int^{\pi/3}_{-\pi/3}\frac{1-\cos2t}{2}dt$$ $$A=\frac{1}{2}\int^{\pi/3}_{-\pi/3}\sec^2tdt+4\Big(\int^{\pi/3}_{-\pi/3}\frac{1}{2}dt-\frac{1}{2}\int^{\pi/3}_{-\pi/3}\cos2tdt\Big)$$ $$A=\frac{1}{2}\tan t\Big]^{\pi/3}_{-\pi/3}+4\Big(\frac{1}{2}t\Big]^{\pi/3}_{-\pi/3}-\frac{1}{4}\sin2t\Big]^{\pi/3}_{-\pi/3}\Big)$$ $$A=\frac{1}{2}\tan t\Big]^{\pi/3}_{-\pi/3}+2t\Big]^{\pi/3}_{-\pi/3}-\sin2t\Big]^{\pi/3}_{-\pi/3}\Big)$$ $$A=\frac{1}{2}(\sqrt3-(-\sqrt3))+2(\pi/3-(-\pi/3))-(\sin(2\pi/3)-\sin(-2\pi/3))$$ $$A=\frac{1}{2}(2\sqrt3)+2(2\pi/3)-(\sqrt3/2-(-\sqrt3/2))$$ $$A=\sqrt3+\frac{4\pi}{3}-\sqrt3$$ $$A=\frac{4\pi}{3}$$