Answer
$$2$$
Work Step by Step
\begin{aligned}
\text{Area}&=\int_{0}^{\pi/2} \cos ^{3} x d x -\int_{\pi/2}^{3\pi/2} \cos ^{3} x d x \\
&=\int_{0}^{\pi/2} \cos ^{2} x \cos x d x -\int_{\pi/2}^{3\pi/2} \cos ^{2} x \cos x d x\\
&=\int_{0}^{\pi/2}\left(1-\sin ^{2} x\right) \cos x d x-\int_{\pi/2}^{3\pi/2}\left(1-\sin ^{2} x\right) \cos x d x \\
&=\int_{0}^{\pi/2}(\cos x d x)-\int_{0}^{\pi/2}\left(\sin ^{2} x\right) \cos x d x-\int_{\pi/2}^{3\pi/2}(\cos x d x)+\int_{\pi/2}^{3\pi/2}\left(\sin ^{2} x\right) \cos x d x \\
&=\sin x-\frac{1}{3} \sin ^{3} x\bigg|_{0}^{\pi/2}-\left(\sin x-\frac{1}{3} \sin ^{3} x\right)\bigg|_{\pi/2}^{3\pi/2}\\
&= 2
\end{aligned}
