Answer
\[ = \frac{\pi }{2} + \frac{1}{6}\]
Work Step by Step
\[\begin{gathered}
\int_0^\pi {\,\left( {1 - {{\cos }^2}3\theta } \right)\,d\theta } \hfill \\
\hfill \\
{\text{using}}\,\,the\,\,identity\,\,{\cos ^2}\,x = \frac{{1 + \cos 2x}}{2} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= \int_0^\pi {\,\left( {1 - \frac{{1 + \cos 6\theta }}{2}} \right)} \,d\theta \hfill \\
\hfill \\
= \int_0^\pi {\,\left( {\frac{1}{2} + \frac{{\cos 6\theta }}{2}} \right)\,d\theta } \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \,\,\left[ {\frac{\theta }{2} + \frac{{\cos \,\,6\,\theta }}{{12}}} \right]_0^\pi \hfill \\
\hfill \\
Fundamental\,\,theorem \hfill \\
\hfill \\
= \,\,\left[ {\frac{\pi }{2} + \frac{{\cos \,6\pi }}{{12}}} \right] - \,\,\left[ {\frac{0}{2} - \frac{{\cos \,0}}{2}} \right] \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \frac{\pi }{2} + \frac{1}{{12}} + \frac{1}{{12}} \hfill \\
\hfill \\
= \frac{\pi }{2} + \frac{1}{6} \hfill \\
\hfill \\
\end{gathered} \]