Calculus: Early Transcendentals (2nd Edition)

$= \frac{{{\pi ^2}}}{2}$
$\begin{gathered} \pi \int_0^\pi {\,{{\left( {\sin x - 0} \right)}^2}\,dx} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ = \pi \int_0^\pi {{{\sin }^2}xdx} \hfill \\ \hfill \\ use\,power\,reducing\,formula \hfill \\ \hfill \\ {\sin ^2}x = \frac{{1 - \cos 2x}}{2} \hfill \\ \hfill \\ = \pi \int_0^\pi {\frac{{\,\left( {1 - \cos 2x} \right)}}{2}\,dx} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ = \frac{\pi }{2}\int_0^\pi {\,\left( {1 - \cos 2x} \right)dx} \hfill \\ \hfill \\ let\,\,u = 2x\,\,\,then\,\,du = 2dx \hfill \\ \hfill \\ and\,\,\,\,\frac{1}{2}du = dx \hfill \\ \hfill \\ = \frac{\pi }{2}\int_0^\pi {\,\left( {1 - \frac{1}{2}\cos u} \right)du} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \frac{\pi }{2}\,\,\left[ {u - \frac{1}{2}\sin u} \right] \hfill \\ \hfill \\ = \frac{\pi }{2}\,\,\left[ {x - \frac{1}{2}\sin 2x} \right]_0^\pi \hfill \\ \hfill \\ evaluate\,the\,\,limits \hfill \\ \hfill \\ = \frac{\pi }{2}\,\,\left[ {\pi - \frac{1}{2}\sin 2\pi - 0} \right] \hfill \\ \hfill \\ {\text{Simplify}} \hfill \\ \hfill \\ = \frac{\pi }{2}\,\,\,\left[ {\pi - 0} \right] = \frac{{{\pi ^2}}}{2} \hfill \\ \end{gathered}$