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

Options

A. `1/2 cos x - 1/10cos 5x +C`
B. `-1/2 sin x +1/10 sin 5x +C`
C. `-1/2cos x + 1/10 cos 5x + C`
D. none of these

Explanation

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

\[ 2 \cos(a) \sin(b) = \sin(a + b) - \sin(a - b) \]

### Step-by-step Solution:

1. **Apply the Identity**:
We can rewrite \( \cos(3x) \sin(2x) \) using the identity.
\[ \cos(3x) \sin(2x) = \frac{1}{2} \left( \sin(3x + 2x) - \sin(3x - 2x) \right) \]
This simplifies to:
\[ \cos(3x) \sin(2x) = \frac{1}{2} \left( \sin(5x) - \sin(x) \right) \]

2. **Set up the Integral**:
Now we can substitute this back into the integral:
\[ \int \cos(3x) \sin(2x) \, dx = \int \frac{1}{2} \left( \sin(5x) - \sin(x) \right) \, dx \]

3. **Factor out the Constant**:
The constant \( \frac{1}{2} \) can be factored out of the integral:
\[ = \frac{1}{2} \int \left( \sin(5x) - \sin(x) \right) \, dx \]

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

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

5. **Simplify the Expression**:
Combining the terms gives:
\[ = -\frac{1}{10} \cos(5x) + \frac{1}{2} \cos(x) + C \]

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