`int (cos 2x)/(sin^(2) x cos^(2)x) dx = ?`

Options

A. `cot x + tan x + C`
B. `-cot x + tan x + C`
C. `cot x - tan x + C`
D. `-cot x - tanx + C`

Explanation

To solve the integral \(\int \frac{\cos 2x}{\sin^2 x \cos^2 x} \, dx\), we can follow these steps:

### Step 1: Rewrite \(\cos 2x\)
We know from trigonometric identities that:
\[ \cos 2x = \cos^2 x - \sin^2 x \]
Thus, we can rewrite the integral as:
\[ \int \frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x} \, dx \]

### Step 2: Split the Integral
We can split the integral into two separate integrals:
\[ \int \frac{\cos^2 x}{\sin^2 x \cos^2 x} \, dx - \int \frac{\sin^2 x}{\sin^2 x \cos^2 x} \, dx \]
This simplifies to:
\[ \int \frac{1}{\sin^2 x} \, dx - \int \frac{1}{\cos^2 x} \, dx \]

### Step 3: Use Trigonometric Identities
We know that:
\[ \frac{1}{\sin^2 x} = \csc^2 x \quad \text{and} \quad \frac{1}{\cos^2 x} = \sec^2 x \]
Thus, we can rewrite the integrals:
\[ \int \csc^2 x \, dx - \int \sec^2 x \, dx \]

### Step 4: Integrate Each Term
The integral of \(\csc^2 x\) is \(-\cot x\) and the integral of \(\sec^2 x\) is \(\tan x\):
\[ -\cot x - \tan x + C \]

### Final Answer
Thus, the solution to the integral is:
\[ \int \frac{\cos 2x}{\sin^2 x \cos^2 x} \, dx = -\cot x - \tan x + C \]