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