Answer
$2$
Work Step by Step
On the interval $[-\displaystyle \pi,-\frac{\pi}{2}]$, the graph of $y=f(x)=\displaystyle \frac{\pi}{2}(\cos x)[\sin(\pi+\pi\sin x)]$ is below the graph of $y=0$,
and on $[-\displaystyle \frac{\pi}{2},0], y=f(x)$ is above $y=0$
The area between the two graphs over [a,b], where one graph is above the other, is given with $\displaystyle \int_{a}^{b}(y_{above}-y_{below})dx$.
$A=\displaystyle \int_{-\pi}^{-\pi/2}[0- f(x)]dx+\int_{-\pi}^{-\pi/2}[f(x)-0]dx$
$=-\displaystyle \int_{-\pi}^{-\pi/2}f(x)dx+\int_{-\pi/2}^{0}f(x)dx$
$A=-I_{1}+I_{2}$
$f(x)=\displaystyle \frac{\pi}{2}(\cos x)[\sin(\pi+\pi\sin x)]$
Apply substitution (in both integrals):
$\left[\begin{array}{lll}
u=\pi+\pi\sin x & & du=\pi\cos xdx\\
& & (\cos x)dx=\frac{du}{\pi}\\
\text{Borders:} & & \\
\text{Integral }I_{1} & & \\
x=-\pi & \rightarrow & u=\pi+\pi(0)=\pi\\
x=-\pi/2 & \rightarrow & u=\pi+\pi(-1)=0\\
\text{Integral }I_{2} & & \\
x=-\pi/2 & \rightarrow & u=0\\
x=0 & \rightarrow & u=\pi+\pi(0)=\pi
\end{array}\right]$
$A=-\displaystyle \int_{\pi}^{0}\frac{\pi}{2}(\sin u)(\frac{1}{\pi}du)+\int_{0}^{\pi}\frac{\pi}{2}(\sin u)(\frac{1}{\pi}du)$
Reverse the borders in the first integral,
$A=\displaystyle \int_{0}^{\pi}\frac{\pi}{2}(\sin u)(\frac{1}{\pi}du)+\int_{0}^{\pi}\frac{\pi}{2}(\sin u)(\frac{1}{\pi}du) =2\cdot\int_{0}^{\pi}\frac{\pi}{2}(\sin u)(\frac{1}{\pi}du)$
$=2\displaystyle \int_{0}^{\pi}(\sin u)(du)$
$=[-\cos u]_{0}^{\pi}$
$=(-\cos\pi)-(-\cos 0)$
$=-(-1)-(-1)$
$=2$
