One way would be to write, via the double angle identity for cosine,
[tex]\dfrac{\cos2x}{\cos^2x\sin^2x}=\dfrac{\cos^2x-\sin^2x}{\cos^2x\sin^2x}=\dfrac1{\sin^2x}-\dfrac1{\cos^2x}=\csc^2x-\sec^2x[/tex]
then recall that [tex](-\cot x)'=\csc^2x[/tex] and [tex](\tan x)'=\sec^2x[/tex].
Alternatively, using the double angle identity for sine,
[tex]\dfrac{\cos2x}{\cos^2x\sin^2x}=\dfrac{\cos2x}{\frac14\sin^22x}[/tex]
which sets up a natural substitution.
[tex]\displaystyle4\int\frac{\cos2x}{\sin^22x}\,\mathrm dx=2\int\frac{\mathrm d(\sin2x)}{\sin^22x}=-\frac2{\sin2x}+C[/tex]
