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

Options

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

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\cos 2x}{\sin^2 x \cos^2 x} \, dx \]
Using the identity for \(\cos 2x\):
\[ \cos 2x = \cos^2 x - \sin^2 x \]
we can rewrite the integral as:
\[ I = \int \frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x} \, dx \]

### Step 2: Split the Integral
Now we can split the integral into two parts:
\[ I = \int \frac{\cos^2 x}{\sin^2 x \cos^2 x} \, dx - \int \frac{\sin^2 x}{\sin^2 x \cos^2 x} \, dx \]
This simplifies to:
\[ I = \int \frac{1}{\sin^2 x} \, dx - \int \frac{1}{\cos^2 x} \, dx \]

### Step 3: Integrate Each Part
Now we can integrate each part separately:
1. The integral of \(\frac{1}{\sin^2 x}\) is \(-\cot x\).
2. The integral of \(\frac{1}{\cos^2 x}\) is \(\tan x\).

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

### Final Result
So the final answer is:
\[ I = -\cot x - \tan x + C \]

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