`int(sin^2x-cos^2x)/(sin^2xcos^2x)dx` is equal to

Options

A. `tanx+cotx+c`
B. `tanx+cose cx+c`
C. `-tanx+cotx+c`
D. `tanx+secx+c`

Explanation

To solve the integral

\[ \int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} \, dx, \]

we will follow these steps:

### Step 1: Rewrite the Integral
We can rewrite the integral as:

\[ \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} - \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) \, dx = \int \left( \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x} \right) \, dx. \]

### Step 2: Simplify the Terms
This simplifies to:

\[ \int \sec^2 x \, dx - \int \csc^2 x \, dx. \]

### Step 3: Integrate Each Term
Now we can integrate each term separately:

1. The integral of \(\sec^2 x\) is \(\tan x\).
2. The integral of \(\csc^2 x\) is \(-\cot x\).

Thus, we have:

\[ \int \sec^2 x \, dx - \int \csc^2 x \, dx = \tan x + \cot x + C, \]

where \(C\) is the constant of integration.

### Final Answer
So, the final result is:

\[ \tan x + \cot x + C. \]