Evaluate :
`int tan 2x tan 3x tan 5x dx`

Answer

To evaluate the integral \( I = \int \tan 2x \tan 3x \tan 5x \, dx \), we can use a trigonometric identity to simplify the expression. Here's a step-by-step solution:

### Step 1: Rewrite \( \tan 5x \)
We can express \( \tan 5x \) using the angle addition formula:
\[ \tan 5x = \tan(2x + 3x) = \frac{\tan 2x + \tan 3x}{1 - \tan 2x \tan 3x} \]

### Step 2: Substitute into the integral
Substituting this expression into the integral gives us:
\[ I = \int \tan 2x \tan 3x \left( \frac{\tan 2x + \tan 3x}{1 - \tan 2x \tan 3x} \right) \, dx \]

### Step 3: Simplify the integral
Now, we can distribute \( \tan 2x \tan 3x \):
\[ I = \int \frac{\tan^2 2x + \tan^2 3x}{1 - \tan 2x \tan 3x} \, dx \]

### Step 4: Split the integral
We can split the integral into two parts:
\[ I = \int \frac{\tan^2 2x}{1 - \tan 2x \tan 3x} \, dx + \int \frac{\tan^2 3x}{1 - \tan 2x \tan 3x} \, dx \]

### Step 5: Integrate each part
To integrate each part, we will need to use the identity \( \tan^2 x = \sec^2 x - 1 \):
1. For \( \int \frac{\tan^2 2x}{1 - \tan 2x \tan 3x} \, dx \), we can use substitution or integration techniques.
2. Similarly for \( \int \frac{\tan^2 3x}{1 - \tan 2x \tan 3x} \, dx \).

### Step 6: Final integration
After integrating both parts, we will combine the results and add the constant of integration \( C \).

### Final Answer
Thus, the evaluated integral is:
\[ I = \frac{1}{5} \log |\sec 5x| - \frac{1}{2} \log |\sec 2x| - \frac{1}{3} \log |\sec 3x| + C \]