`int 1/(sin^2x cos^2x )dx` is equal to:

Options

A. `tanx+cotx+c`
B. `tanx-cotx+c`
C. `tanx.cotx+c`
D. `tanx-cot2x+c`

Answer

To solve the integral \( \int \frac{1}{\sin^2 x \cos^2 x} \, dx \), we can start by rewriting the integrand using trigonometric identities.

### Step 1: Rewrite the integrand
We know that:
\[ \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x) \]
Thus, we can rewrite the integral as:
\[ \int \frac{1}{\sin^2 x \cos^2 x} \, dx = \int \frac{4}{\sin^2(2x)} \, dx \]

### Step 2: Use the cosecant function
The integral can now be expressed in terms of the cosecant function:
\[ \int \frac{4}{\sin^2(2x)} \, dx = 4 \int \csc^2(2x) \, dx \]

### Step 3: Integrate using the known integral
We know that:
\[ \int \csc^2(u) \, du = -\cot(u) + C \]
In our case, \( u = 2x \), so \( du = 2 \, dx \) or \( dx = \frac{du}{2} \). Therefore, we have:
\[ 4 \int \csc^2(2x) \, dx = 4 \cdot \frac{1}{2} \int \csc^2(u) \, du = 2 \int \csc^2(u) \, du \]
Substituting back:
\[ = 2 \left( -\cot(2x) + C \right) = -2 \cot(2x) + C \]

### Final Answer
Thus, the solution to the integral is:
\[ \int \frac{1}{\sin^2 x \cos^2 x} \, dx = -2 \cot(2x) + C \]