Evaluate `int tan^(3)x dx`

Explanation

To evaluate the integral \( \int \tan^3 x \, dx \), we can follow these steps:

### Step 1: Rewrite \( \tan^3 x \)
We can express \( \tan^3 x \) as:
\[ \tan^3 x = \tan x \cdot \tan^2 x \]
Using the identity \( \tan^2 x = \sec^2 x - 1 \), we can rewrite it as:
\[ \tan^3 x = \tan x \cdot (\sec^2 x - 1) \]

### Step 2: Substitute into the integral
Now, substitute this back into the integral:
\[ \int \tan^3 x \, dx = \int \tan x (\sec^2 x - 1) \, dx \]
This can be separated into two integrals:
\[ \int \tan x \sec^2 x \, dx - \int \tan x \, dx \]

### Step 3: Evaluate \( \int \tan x \sec^2 x \, dx \)
For the first integral, we can use the substitution \( u = \tan x \), which gives \( du = \sec^2 x \, dx \):
\[ \int \tan x \sec^2 x \, dx = \int u \, du = \frac{u^2}{2} + C = \frac{\tan^2 x}{2} + C \]

### Step 4: Evaluate \( \int \tan x \, dx \)
The second integral can be evaluated using the known result:
\[ \int \tan x \, dx = -\ln |\cos x| + C \]

### Step 5: Combine the results
Now we combine both results:
\[ \int \tan^3 x \, dx = \frac{\tan^2 x}{2} - (-\ln |\cos x|) + C \]
This simplifies to:
\[ \int \tan^3 x \, dx = \frac{\tan^2 x}{2} + \ln |\cos x| + C \]

### Final Answer
Thus, the final result is:
\[ \int \tan^3 x \, dx = \frac{\tan^2 x}{2} + \ln |\cos x| + C \]