`int(1)/(sin^(2)x cos^(2)x) dx = ?`

Options

A. `tanx + cot x + C`
B. `-tan x + cot x + C`
C. `tanx - cot x + C`
D. none of these

Explanation

To solve the integral

\[ \int \frac{1}{\sin^2 x \cos^2 x} \, dx, \]

we can start by rewriting the integrand using the identity \( \sin^2 x + \cos^2 x = 1 \).

### Step 1: Rewrite the integrand
We can express the integrand as follows:

\[ \frac{1}{\sin^2 x \cos^2 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x}. \]

### Step 2: Split the fraction
Now, we can split the fraction into two parts:

\[ \frac{1}{\sin^2 x \cos^2 x} = \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x}. \]

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

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

### Step 4: Use known integrals
We know that:

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

### Step 5: Combine the results
Thus, we can combine the results of the integrals:

\[ \int \frac{1}{\sin^2 x \cos^2 x} \, dx = \tan x - \cot x + C. \]

### Final Answer
Therefore, the final answer is:

\[ \int \frac{1}{\sin^2 x \cos^2 x} \, dx = \tan x - \cot x + C. \]

---