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

Answer

To solve the integral \( \int \frac{1}{\sin x \cos x + 2 \cos^2 x} \, dx \), we will follow these steps:

### Step 1: Simplify the Denominator
We start by rewriting the denominator:
\[ \sin x \cos x + 2 \cos^2 x = \cos^2 x \left( \frac{\sin x}{\cos^2 x} + 2 \right) = \cos^2 x \left( \tan x + 2 \right) \]
Thus, we can rewrite the integral as:
\[ \int \frac{1}{\cos^2 x (\tan x + 2)} \, dx \]

### Step 2: Rewrite the Integral
We can express \( \frac{1}{\cos^2 x} \) as \( \sec^2 x \):
\[ \int \frac{\sec^2 x}{\tan x + 2} \, dx \]

### Step 3: Substitution
Let \( t = \tan x \). Then, we have:
\[ dt = \sec^2 x \, dx \quad \Rightarrow \quad dx = \frac{dt}{\sec^2 x} \]
Substituting this into the integral gives:
\[ \int \frac{1}{t + 2} \, dt \]

### Step 4: Integrate
The integral \( \int \frac{1}{t + 2} \, dt \) can be solved as:
\[ \ln |t + 2| + C \]

### Step 5: Back Substitute
Now, we substitute back \( t = \tan x \):
\[ \ln |\tan x + 2| + C \]

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