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

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{1}{\sin x \cos x + 2 \cos^2 x} \, dx \]
We can factor out \(\cos^2 x\) from the denominator:
\[ I = \int \frac{1}{\cos^2 x \left(\frac{\sin x}{\cos x} + 2\right)} \, dx \]
This simplifies to:
\[ I = \int \frac{1}{\cos^2 x \left(\tan x + 2\right)} \, dx \]

### Step 2: Substitute for Secant
Since \(\frac{1}{\cos^2 x} = \sec^2 x\), we rewrite the integral:
\[ I = \int \frac{\sec^2 x}{\tan x + 2} \, dx \]

### Step 3: Substitution
Let \( t = \tan x \). Then, the derivative \( dt = \sec^2 x \, dx \). Thus, we can rewrite the integral in terms of \( t \):
\[ I = \int \frac{1}{t + 2} \, dt \]

### Step 4: Integrate
The integral of \(\frac{1}{t + 2}\) is:
\[ I = \ln |t + 2| + C \]

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

### Final Answer
Thus, the final result of the integral is:
\[ I = \ln |\tan x + 2| + C \]

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