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

Options

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

Answer

To solve the integral

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

we can start by rewriting the integrand.

### Step 1: Rewrite the integrand

We can separate the integral into two parts:

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

This simplifies to:

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

### Step 2: Use trigonometric identities

Recall that:

\[ \frac{1}{\cos^2 x} = \sec^2 x \quad \text{and} \quad \frac{1}{\sin^2 x} = \csc^2 x. \]

Thus, we can rewrite the integral as:

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

### 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 \(-\cot x\).

So we have:

\[ I = \tan x - (-\cot x) + C, \]

which simplifies to:

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

### Final Result

Thus, the integral

\[ \int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} \, dx = \tan x + \cot x + C. \]

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