Answer
$\displaystyle \frac{\pi}{2}$
Work Step by Step
On the interval $[0,\pi]$, the graph of $y=\cos^{2}x$ is below the graph of $y=1$.
On the interval $[0,\pi]$, the graph of $y=\cos^{2}x$ is below the graph of $y=1$.
The area between the two graphs over [a,b], where one graph is above the other, is given with $\displaystyle \int_{a}^{b}(y_{above}-y_{below})dx$.
Thus,
$A=\displaystyle \int_{0}^{\pi}(1-\cos^{2}x)dx=\int_{0}^{\pi}(\sin^{2}x)dx$
Using the double angle formula for $\cos 2x$
$\cos 2x=\cos^{2}x-\sin^{2}x=-(1-\sin^{2}x)-\sin^{2}x=1-2\sin^{2}x$,
We transform $\displaystyle \sin^{2}x= \frac{1}{2}(1-\cos 2x)$
$A= \displaystyle \frac{1}{2}\int_{0}^{\pi}(1-\cos 2x)dx= \displaystyle \frac{1}{2}\int_{0}^{\pi}dx- \displaystyle \frac{1}{2}\int_{0}^{\pi}\cos 2xdx$
For the second integral:
$\left[\begin{array}{lll}
u=2x, & & du=2dx\\
& & dx=du/2\\
\text{Borders:} & & \\
x=0 & \rightarrow & u=2(0)=0\\
x=\pi & \rightarrow & u=2\pi
\end{array}\right]$
$= \displaystyle \frac{1}{2}[x]_{0}^{\pi}- \displaystyle \frac{1}{2}\int_{0}^{2\pi}\cos u(\frac{du}{2})$
$=\displaystyle \frac{1}{2}(\pi-0)-\frac{1}{4}[\sin u]_{0}^{2\pi}$
$= \displaystyle \frac{\pi}{2}-\frac{1}{4}(0-0)$
= $\displaystyle \frac{\pi}{2}$
