Answer
$2^{5/2}$
Work Step by Step
On the interval $[-\pi,0]$, the graph of $y=3(\sin x)\sqrt{1+\cos x}$ is below the graph of $y=0$.
The area between the two graphs over [a,b], where one graph is above the other, is given with $\displaystyle \int_{a}^{b}(y_{above}-y_{below})dx$.
$A=\displaystyle \int_{-\pi}^{0}[0-3(\sin x)\sqrt{1+\cos x}]dx=-3\int_{-\pi}^{0}\sqrt{1+\cos x}(\sin x)dx=$
Apply substitution:
$\left[\begin{array}{lll}
u=1+\cos x & & du=-\sin xdx\\
& & (\sin x)dx=-du\\
Borders: & & \\
x=-\pi & \rightarrow & u=1+\cos(-\pi)=0\\
x=0 & \rightarrow & u=1+\cos(0)=2
\end{array}\right]$
$=-\displaystyle \int_{0}^{2}3u^{1/2}(-du)$
$=3\displaystyle \int_{0}^{2}u^{1/2}du$
$=[2u^{3/2}]_{0}^{2}$
$=2(2)^{3/2}-0$
= $2^{5/2}$
