Evaluate: `int (tan x-x)tan^(2)xdx`

Answer

To evaluate the integral \( \int (\tan x - x) \tan^2 x \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ \int (\tan x - x) \tan^2 x \, dx \]
We know that \( \tan^2 x = \sec^2 x - 1 \). Thus, we can rewrite the integral as:
\[ \int (\tan x - x)(\sec^2 x - 1) \, dx \]

### Step 2: Distribute the terms
Now we distribute \( (\tan x - x) \) across \( (\sec^2 x - 1) \):
\[ \int (\tan x - x) \sec^2 x \, dx - \int (\tan x - x) \, dx \]

### Step 3: Evaluate the first integral
For the first integral \( \int (\tan x - x) \sec^2 x \, dx \), we can use substitution. Let:
\[ t = \tan x - x \]
Then, differentiating gives:
\[ dt = \sec^2 x \, dx - dx = (\sec^2 x - 1) \, dx \]
This implies:
\[ dx = \frac{dt}{\sec^2 x - 1} \]
However, we can also note that \( \sec^2 x = 1 + \tan^2 x \).

### Step 4: Evaluate the second integral
Now, we evaluate the second integral:
\[ \int (\tan x - x) \, dx \]
This can be split into two separate integrals:
\[ \int \tan x \, dx - \int x \, dx \]
The integral of \( \tan x \) is:
\[ -\ln |\cos x| + C \]
And the integral of \( x \) is:
\[ \frac{x^2}{2} + C \]

### Step 5: Combine results
Now we combine our results:
\[ \int (\tan x - x) \tan^2 x \, dx = \int (\tan x - x) \sec^2 x \, dx - \left( -\ln |\cos x| + \frac{x^2}{2} \right) \]

### Step 6: Final expression
Thus, the final expression for the integral is:
\[ \int (\tan x - x) \tan^2 x \, dx = \frac{(tan x - x)^2}{2} + C \]