`int tan x tan 2x tan 3x dx `

Options

A. `-1/3 "log sin 3x" + 1/2 "log sin 2x" +"log sin x"`
B. `-1/3 "log cos 3x" + 1/2 "log cos 2x " + "log cos x"`
C. `-1/3 "log sin 3x" + 1/2 "log cos 2x " + "log sin x"`
D. `-1/3 "log cos 3x" + 1/2 "log sin 2x " + "log cos x"`

Answer

To solve the integral \( \int \tan x \tan 2x \tan 3x \, dx \), we can use the properties of trigonometric functions and some identities. Here’s a step-by-step solution:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int \tan x \tan 2x \tan 3x \, dx \]

### Step 2: Use the identity for tangent
Recall that:
\[ \tan A = \frac{\sin A}{\cos A} \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{\sin x}{\cos x} \cdot \frac{\sin 2x}{\cos 2x} \cdot \frac{\sin 3x}{\cos 3x} \, dx \]

### Step 3: Expand the product
Now, we can express the product of sines and cosines:
\[ I = \int \frac{\sin x \sin 2x \sin 3x}{\cos x \cos 2x \cos 3x} \, dx \]

### Step 4: Use the product-to-sum identities
We can apply the product-to-sum identities for sine functions:
\[ \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \]
Using this identity, we can simplify \( \sin x \sin 2x \) and then multiply by \( \sin 3x \).

### Step 5: Simplify the integral
After applying the product-to-sum identities, we will have a more manageable integral. Let’s denote the resulting integral as \( J \):
\[ J = \int f(x) \, dx \]
where \( f(x) \) is the simplified function obtained from the previous step.

### Step 6: Solve the integral
Now we can integrate \( J \) using standard integration techniques. This may involve further simplification or substitution.

### Step 7: Combine results
Finally, we will combine the results of our integration and apply any necessary constants of integration.

### Final Result
The final result will be:
\[ I = \text{(result from integrating } J\text{)} + C \]
where \( C \) is the constant of integration.