## Thomas' Calculus 13th Edition

$$0$$
\eqalign{ & \int_{ - \pi /2}^{\pi /2} {\cos x\cos 7x} dx \cr & {\text{Use the identity }}\cos mx\cos xnx = \frac{1}{2}\left[ {\cos \left( {m - n} \right)x + \cos \left( {m + n} \right)x} \right]\,\,\left( {{\text{see page 463}}} \right) \cr & \int_{ - \pi /2}^{\pi /2} {\cos x\cos 7x} dx = \int_{ - \pi /2}^{\pi /2} {\frac{1}{2}\left[ {\cos \left( {1 - 7} \right)x + \cos \left( {1 + 7} \right)x} \right]dx} \cr & = \frac{1}{2}\int_{ - \pi /2}^{\pi /2} {\left( {\cos 6x + \cos 8x} \right)} \cr & {\text{integrating}} \cr & = \frac{1}{2}\left( {\frac{1}{6}\sin 6x + \frac{1}{8}\sin 8x} \right)_{ - \pi /2}^{\pi /2} \cr & = \frac{1}{2}\left( {\frac{1}{6}\sin 6\left( {\frac{\pi }{2}} \right) + \frac{1}{8}\sin 8\left( {\frac{\pi }{2}} \right)} \right) - \frac{1}{2}\left( {\frac{1}{6}\sin 6\left( { - \frac{\pi }{2}} \right) + \frac{1}{8}\sin 8\left( { - \frac{\pi }{2}} \right)} \right) \cr & {\text{simplifying}} \cr & = \frac{1}{2}\left( {\frac{1}{6}\sin \left( {3\pi } \right) + \frac{1}{8}\sin \left( {4\pi } \right)} \right) - \frac{1}{2}\left( {\frac{1}{6}\sin \left( { - 3\pi } \right) + \frac{1}{8}\sin 8\left( { - 4\pi } \right)} \right) \cr & = \frac{1}{2}\left( 0 \right) - \frac{1}{2}\left( 0 \right) \cr & = 0 \cr}