Thomas' Calculus 13th Edition

$\displaystyle \frac{\pi}{2}$
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: $\left[\begin{array}{lll} u=2x, & & du=2dx\\ & & dx=du/2\\ \text{Borders:} & & \\ x=0 & \rightarrow & u=2(0)=0\\ x=\pi & \rightarrow & u=2\pi \end{array}\right]$ $= \displaystyle \frac{1}{2}[x]_{0}^{\pi}- \displaystyle \frac{1}{2}\int_{0}^{2\pi}\cos u(\frac{du}{2})$ $=\displaystyle \frac{1}{2}(\pi-0)-\frac{1}{4}[\sin u]_{0}^{2\pi}$ $= \displaystyle \frac{\pi}{2}-\frac{1}{4}(0-0)$ = $\displaystyle \frac{\pi}{2}$