## Thomas' Calculus 13th Edition

$$\frac{\pi }{8}$$
\eqalign{ & \int_0^{\pi /4} {{{\cos }^2}\left( {4t - \frac{\pi }{4}} \right)} dt \cr & {\text{Use co}}{{\text{s}}^2}\theta = \frac{1}{2} + \frac{1}{2}\cos 2\theta \cr & = \int_0^{\pi /4} {\left( {\frac{1}{2} + \frac{1}{2}\cos 2\left( {4t - \frac{\pi }{4}} \right)} \right)} dt \cr & = \int_0^{\pi /4} {\left( {\frac{1}{2} + \frac{1}{2}\cos \left( {8t - \frac{\pi }{2}} \right)} \right)} dt \cr & {\text{Use the cofunction identity} }\cr & \left( {cos \theta - \frac{\pi }{2}} \right) = \sin \theta \cr & = \int_0^{\pi /4} {\left( {\frac{1}{2} + \frac{1}{2}\sin \left( {8t} \right)} \right)} dt \cr & {\text{Integrating}} \cr & = \left( {\frac{1}{2}t - \frac{1}{{16}}\cos 8t} \right)_0^{\pi /4} \cr & {\text{Evaluating}} \cr & = \left( {\frac{1}{2}\left( {\frac{\pi }{4}} \right) - \frac{1}{{16}}\cos 8\left( {\frac{\pi }{4}} \right)} \right) - \left( {\frac{1}{2}\left( 0 \right) - \frac{1}{{16}}\cos 8\left( 0 \right)} \right) \cr & {\text{Simplifying}} \cr & = \frac{\pi }{8} - \frac{1}{{16}}\left( 1 \right) + \frac{1}{{16}} \cr & = \frac{\pi }{8} \cr}