`int(cos2x)/(sin^(2)x.cos^(2)x)dx=`

Options

A. `tanx-cotx`
B. `cotx-tanx`
C. `tanx+cotx`
D. `-(tanx+cotx)`

Answer

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

### Step 1: Rewrite \( \cos 2x \)
Using the trigonometric identity for cosine, we know 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: Separate the Integral
Now, we can separate the integral into two parts:
\[ \int \frac{\cos^2 x}{\sin^2 x \cos^2 x} \, dx - \int \frac{\sin^2 x}{\sin^2 x \cos^2 x} \, dx \]
This simplifies to:
\[ \int \frac{1}{\sin^2 x} \, dx - \int \frac{1}{\cos^2 x} \, dx \]

### Step 3: Rewrite the Integrals
The integrals can be rewritten using the identities for cosecant and secant:
\[ \int \csc^2 x \, dx - \int \sec^2 x \, dx \]

### Step 4: Integrate
Now we can integrate each term:
1. The integral of \( \csc^2 x \) is \( -\cot x \).
2. The integral of \( \sec^2 x \) is \( \tan x \).

Thus, we have:
\[ -\cot x - \tan x + C \]
where \( C \) is the constant of integration.

### Final Solution
The final result of the integral is:
\[ \int \frac{\cos 2x}{\sin^2 x \cos^2 x} \, dx = -\cot x - \tan x + C \]

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