Integrate 1 over sin²xcos²x

To integrate $\frac{1}{\sin^2 x \cos^2 x}$, we can use trigonometric identities.

Step 1: Rewrite the numerator using the identity $\sin^2 x + \cos^2 x = 1$. $\int \frac{1}{\sin^2 x \cos^2 x} dx = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} dx$

Step 2: Split the fraction into two terms. $\int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) dx$

Step 3: Simplify each term. $\int \left( \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \right) dx$ Using the reciprocal identities $\sec^2 x = \frac{1}{\cos^2 x}$ and $\csc^2 x = \frac{1}{\sin^2 x}$: $\int (\sec^2 x + \csc^2 x) dx$

Step 4: Integrate each term. We know that $\int \sec^2 x \, dx = \tan x + C$ and $\int \csc^2 x \, dx = -\cot x + C$. $\int (\sec^2 x + \csc^2 x) dx = \tan x - \cot x + C$

The final answer is $\boxed{\tan x - \cot x + C}$.