$\int \frac{dx}{\cos (x+4)\cos (x+2)}=$

$\begin{array}{l}\int \frac{dx}{\cos (x+4)\cos (x+2)}=\frac{1}{\sin 2}\int \frac{\sin \left[\right(x+4)-(x+2\left)\right]}{\cos (x+4)\cos (x+2)}dx\\ =\frac{1}{\sin 2}\int \frac{\sin (x+4)\cos (x+2)-\cos (x+4)\sin (x+2)}{\cos (x+4)\cos (x+2)}dx=\frac{1}{\sin 2}\int [\tan (x+4)-\tan (x+2\left)\right]dx\\ =\frac{1}{\sin 2}[\log |\sec (x+4)|-\log |\sec (x+2)\left|\right]+c=\frac{1}{\sin 2}\log \left|\frac{\sec (x+4)}{\sec (x+2)}\right|+c.\end{array}$