`int sin 3x sin 2x dx = ?`

Options

A. `- 1/5 cos 5x + C`
B. `1/2 sin x + 1/10 sin 5x - C`
C. `1/2 sinx - 1/10 sin 5x -+C`
D. `-1/3 cos 3 x - 1/2 sin 2 x + C`

Explanation

To solve the integral \( \int \sin(3x) \sin(2x) \, dx \), we can use the product-to-sum identities for sine functions. The relevant identity is:

\[ \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \]

### Step-by-step Solution:

1. **Identify A and B**:
Here, let \( A = 3x \) and \( B = 2x \).

2. **Apply the Product-to-Sum Identity**:
Using the identity, we can rewrite the integral:
\[ \sin(3x) \sin(2x) = \frac{1}{2} [\cos(3x - 2x) - \cos(3x + 2x)] \]
Simplifying the angles gives:
\[ \sin(3x) \sin(2x) = \frac{1}{2} [\cos(x) - \cos(5x)] \]

3. **Set Up the Integral**:
Substitute this back into the integral:
\[ \int \sin(3x) \sin(2x) \, dx = \int \frac{1}{2} [\cos(x) - \cos(5x)] \, dx \]

4. **Factor Out the Constant**:
The integral can be simplified by factoring out the constant \( \frac{1}{2} \):
\[ = \frac{1}{2} \int [\cos(x) - \cos(5x)] \, dx \]

5. **Integrate Each Term**:
Now, we can integrate each term separately:
- The integral of \( \cos(x) \) is \( \sin(x) \).
- The integral of \( \cos(5x) \) is \( \frac{1}{5} \sin(5x) \).

Thus, we have:
\[ = \frac{1}{2} \left[ \sin(x) - \frac{1}{5} \sin(5x) \right] + C \]

6. **Combine the Results**:
Finally, we can simplify this to:
\[ = \frac{1}{2} \sin(x) - \frac{1}{10} \sin(5x) + C \]

### Final Answer:
\[ \int \sin(3x) \sin(2x) \, dx = \frac{1}{2} \sin(x) - \frac{1}{10} \sin(5x) + C \]