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