`int(1)/(cos x . cot x ) dx`

Answer

To solve the integral \( \int \frac{1}{\cos x \cdot \cot x} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We start with the integral:
\[ \int \frac{1}{\cos x \cdot \cot x} \, dx \]
We know that \( \cot x = \frac{\cos x}{\sin x} \). Therefore, we can rewrite the integrand:
\[ \cot x = \frac{\cos x}{\sin x} \implies \frac{1}{\cot x} = \frac{\sin x}{\cos x} \]
Substituting this into the integral gives:
\[ \int \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} \, dx = \int \frac{\sin x}{\cos^2 x} \, dx \]

### Step 2: Use a trigonometric identity
Now, we can express \( \frac{\sin x}{\cos^2 x} \) in terms of \( \sec x \) and \( \tan x \):
\[ \frac{\sin x}{\cos^2 x} = \sin x \cdot \sec^2 x \]
Thus, we can rewrite the integral as:
\[ \int \sin x \cdot \sec^2 x \, dx \]

### Step 3: Use substitution
Let \( u = \cos x \). Then, the derivative \( du = -\sin x \, dx \) or \( -du = \sin x \, dx \). This transforms our integral:
\[ \int \sin x \cdot \sec^2 x \, dx = -\int \sec^2 x \, du \]
Since \( \sec^2 x = \frac{1}{\cos^2 x} = \frac{1}{u^2} \), we have:
\[ -\int \frac{1}{u^2} \, du \]

### Step 4: Integrate
The integral \( -\int \frac{1}{u^2} \, du \) can be computed as follows:
\[ -\int u^{-2} \, du = -\left(-\frac{1}{u}\right) = \frac{1}{u} + C \]
Substituting back \( u = \cos x \):
\[ \frac{1}{\cos x} + C = \sec x + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{1}{\cos x \cdot \cot x} \, dx = \sec x + C \]