Answer
\[ = \pi \]
Work Step by Step
\[\begin{gathered}
\int_{ - \pi }^\pi {{{\cos }^2}xdx} \hfill \\
\hfill \\
where\,\,{\cos ^2}x = \frac{{1 + \cos 2x}}{2} \hfill \\
\hfill \\
= \int_{ - \pi }^\pi {\,\left( {\frac{{1 + \cos 2x}}{2}} \right)} \,dx \hfill \\
\hfill \\
integrate\,\,and\,\,use\,\,the\,\,ftc \hfill \\
\hfill \\
= \,\left( {\frac{\pi }{2} + \frac{{\sin \,\left( {2\pi } \right)}}{4}} \right) - \,\left( { - \frac{\pi }{2} + \frac{{\sin \,\left( { - 2\pi } \right)}}{4}} \right) \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \frac{\pi }{2} + \frac{\pi }{2} \hfill \\
\hfill \\
= \pi \hfill \\
\end{gathered} \]
