\(\int \frac{{\sin }^{2}x−{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx\text{ is equal to : }\)

Given integral: \(I=\int \frac{(\sin ²x-\cos ²x)}{(\sin ²x\cos ²x)}dx\)

Using the identity: \(\sin ²x-\cos ²x=-\cos 2x\)

The integral becomes: \(I=\int \frac{(-\cos 2x)}{(\sin ²x\cos ²x)}dx\)

Using the identity:\(\sin ²x\cos ²x=(\frac{1}{4})\sin ²2x\)

The integral transforms into: \(I=-4\int \frac{\cos 2x}{\sin ²2x}dx\)

Substituting t=sin2x, so that \(dt=2\cos 2xdx\)

Thus, \(dx=\frac{dt}{(2\cos 2x)}\)

Substituting in the integral:

\(I=-4\int (\frac{\cos 2x}{t²})\times (\frac{dt}{(2\cos 2x)})\\ =-4\int (\frac{dt}{(2t²)})\\ =-2\int t⁻²dt\\ =\frac{2}{t}+c\)

Using the identity we get: \(\frac{1}{\sin x\cos x}=\tan x+cotx\)+ c