Answer
\[ = \frac{\pi }{8}\]
Work Step by Step
\[\begin{gathered}
\int_0^{\frac{\pi }{4}} {{{\cos }^2}8\theta d\theta } \hfill \\
\hfill \\
use\,\,the\,identity\,\,{\cos ^2}x = \frac{{1 + \cos 2x}}{2} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= \int_0^{\frac{\pi }{4}} {\frac{{1 + \cos 16\theta }}{2}} d\theta \hfill \\
\hfill \\
= \int_0^{\frac{\pi }{4}} {\frac{1}{2}d\theta + \frac{1}{{32}}\int_0^{\frac{\pi }{4}} {\cos 16\theta \,\left( {16} \right)d\theta } } \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \,\,\left[ {\frac{\theta }{2}} \right]_0^{\frac{\pi }{4}} + \frac{1}{{32}}\,\,\left[ {\sin 16\theta } \right]_0^{\frac{\pi }{4}} \hfill \\
\hfill \\
Fundamental\,\,theorem \hfill \\
\hfill \\
= \frac{\pi }{8} - 0 + \frac{1}{{32}}\,\,\left[ {\sin \,\left( {\frac{{16\pi }}{4}} \right) - \sin \theta } \right] \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
= \frac{\pi }{8} \hfill \\
\hfill \\
\hfill \\
\end{gathered} \]
