`int(tan x + cos x)^(2) dx `

Answer

To solve the integral \( \int (\tan x + \cot x)^2 \, dx \), we will follow these steps:

### Step 1: Expand the integrand
Using the identity \( (a + b)^2 = a^2 + b^2 + 2ab \), we can expand \( (\tan x + \cot x)^2 \):
\[ (\tan x + \cot x)^2 = \tan^2 x + \cot^2 x + 2 \tan x \cot x \]

### Step 2: Simplify the expression
Recall that \( \tan x \cot x = 1 \). Therefore, we can simplify:
\[ 2 \tan x \cot x = 2 \]
So, we have:
\[ \tan^2 x + \cot^2 x + 2 \]

### Step 3: Rewrite the integral
Now, we can rewrite the integral:
\[ \int (\tan^2 x + \cot^2 x + 2) \, dx = \int \tan^2 x \, dx + \int \cot^2 x \, dx + \int 2 \, dx \]

### Step 4: Integrate each term
1. **Integrate \( \tan^2 x \)**:
\[ \tan^2 x = \sec^2 x - 1 \implies \int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \int \sec^2 x \, dx - \int 1 \, dx \]
The integral of \( \sec^2 x \) is \( \tan x \), and the integral of 1 is \( x \):
\[ \int \tan^2 x \, dx = \tan x - x \]

2. **Integrate \( \cot^2 x \)**:
\[ \cot^2 x = \csc^2 x - 1 \implies \int \cot^2 x \, dx = \int (\csc^2 x - 1) \, dx = \int \csc^2 x \, dx - \int 1 \, dx \]
The integral of \( \csc^2 x \) is \( -\cot x \):
\[ \int \cot^2 x \, dx = -\cot x - x \]

3. **Integrate \( 2 \)**:
\[ \int 2 \, dx = 2x \]

### Step 5: Combine the results
Now, we combine all the integrals:
\[ \int (\tan^2 x + \cot^2 x + 2) \, dx = (\tan x - x) + (-\cot x - x) + 2x \]
Simplifying this gives:
\[ \tan x - \cot x + 0 \]

### Final Answer
Thus, the final result of the integral is:
\[ \int (\tan x + \cot x)^2 \, dx = \tan x - \cot x + C \]