Answer
$$\frac{1}{2}\sin x + \frac{1}{{14}}\sin 7x + C $$
Work Step by Step
$$\eqalign{
& \int {\cos 3x} \cos 4xdx \cr
& {\text{Use the identity }}\cos mx\cos xnx = \frac{1}{2}\left[ {\cos \left( {m - n} \right)x + \cos \left( {m + n} \right)x} \right]\,\,\left( {{\text{see page 463}}} \right) \cr
& \int {\cos 3x} \cos 4xdx = \int {\frac{1}{2}} \left[ {\cos \left( {3 - 4} \right)x + \cos \left( {3 + 4} \right)x} \right]dx \cr
& = \frac{1}{2}\int {\left( {\cos x + \cos 7x} \right)} dx \cr
& {\text{distribute}} \cr
& = \frac{1}{2}\int {\cos x} dx + \frac{1}{2}\int {\cos 7x} dx \cr
& {\text{integrating}} \cr
& = \frac{1}{2}\sin x + \frac{1}{2}\left( {\frac{1}{7}\sin 7x} \right) + C \cr
& = \frac{1}{2}\sin x + \frac{1}{{14}}\sin 7x + C \cr} $$
