$\begin{aligned} & \text { 73. Let } I=\int \frac{d x}{\cos (x+4) \cos (x+2)} \\ & =\frac{1}{\sin 2} \int \frac{\sin 2}{\cos (x+4) \cos (x+2)} d x \\ & =\frac{1}{\sin 2} \int \frac{\sin [(x+4)-(x+2)]}{\cos (x+4) \cos (x+2)} d x=\frac{1}{\sin 2} \\ & \int \frac{[\sin (x+4) \cos (x+2)-\cos (x+4) \sin (x+2)]}{\cos (x+4) \cos (x+2)} d x \\ & =\frac{1}{\sin 2} \int\left[\frac{\sin (x+4)}{\cos (x+4)}-\frac{\sin (x+2)}{\cos (x+2)}\right] d x \\ & =\frac{1}{\sin 2} \int[\tan (x+4)-\tan (x+2)] d x\end{aligned}$
$\begin{aligned} & =\frac{1}{\sin 2} \int \tan (x+4) d x-\frac{1}{\sin 2} \int \tan (x+2) d x \\ & =\frac{1}{\sin 2}[\log \sec (x+4)-\log \sec (x+2)]+C \\ & =\frac{1}{\sin 2} \log \left|\frac{\sec (x+4)}{\sec (x+2)}\right|+C\end{aligned}$