Evaluate : `int tan x tan 2 x tan 3x dx`

Answer

To evaluate the integral \( \int \tan x \tan 2x \tan 3x \, dx \), we can use the identity for the tangent of a sum.

### Step-by-Step Solution:

1. **Use the Identity for \( \tan 3x \)**:
We know that:
\[ \tan 3x = \frac{\tan x + \tan 2x}{1 - \tan x \tan 2x} \]
This can be rearranged to express \( \tan 3x \) in terms of \( \tan x \) and \( \tan 2x \).

2. **Rearranging the Equation**:
From the identity:
\[ \tan 3x - \tan x \tan 2x = \tan x + \tan 2x \]
We can express \( \tan x \tan 2x \tan 3x \) as:
\[ \tan x \tan 2x \tan 3x = \tan 3x - \tan x - \tan 2x \]
This is our first equation.

3. **Substituting into the Integral**:
Now substitute this expression into the integral:
\[ \int \tan x \tan 2x \tan 3x \, dx = \int (\tan 3x - \tan x - \tan 2x) \, dx \]

4. **Integrating Each Term**:
Now we can integrate each term separately:
- The integral of \( \tan 3x \):
\[ \int \tan 3x \, dx = -\frac{1}{3} \log |\cos 3x| + C_1 \]
- The integral of \( \tan x \):
\[ \int \tan x \, dx = -\log |\cos x| + C_2 \]
- The integral of \( \tan 2x \):
\[ \int \tan 2x \, dx = -\frac{1}{2} \log |\cos 2x| + C_3 \]

5. **Combining the Results**:
Now combine all these results:
\[ \int \tan x \tan 2x \tan 3x \, dx = \left(-\frac{1}{3} \log |\cos 3x| + \log |\cos x| + \frac{1}{2} \log |\cos 2x|\right) + C \]
where \( C \) is a constant of integration.

### Final Result:
Thus, the final result is:
\[ \int \tan x \tan 2x \tan 3x \, dx = -\frac{1}{3} \log |\cos 3x| + \log |\cos x| + \frac{1}{2} \log |\cos 2x| + C \]