Question

Integrate cos2x/(cos^2x*sin^2x)

Answer 1

One way would be to write, via the double angle identity for cosine,


`(\cos2x)/(\cos^2x\sin^2x)=(\cos^2x-\sin^2x)/(\cos^2x\sin^2x)=\frac1{\sin^2x}-\frac1{\cos^2x}=\csc^2x-\sec^2x`

then recall that
`(-\cot x)'=\csc^2x` and
`(\tan x)'=\sec^2x`.

Alternatively, using the double angle identity for sine,


`(\cos2x)/(\cos^2x\sin^2x)=(\cos2x)/(\frac14\sin^22x)`

which sets up a natural substitution.


`\displaystyle4\int(\cos2x)/(\sin^22x)\,\mathrm dx=2\int(\mathrm d(\sin2x))/(\sin^22x)=-\frac2{\sin2x}+C`