Solve integral of [(cos ² 2x) / cos ² x ] dx

hi, {cos2x/cos(x)}^2 = {(2cos^2(x)-1)/cosx}^2 => {2 cosx-secx}^2 = {4(cosx)^2+(secx)^2-4} => {2{2(cosx)^2 -1} -2 +(secx^2)} => 2cos(2x)dx -2dx +9secx)^2dx => integral of {cos2x/cos(x)}^2 = integral of 2cos(2x)dx -2dx +9secx)^2dx = sin(2x) -2x +tan(x)+c.