`int("sin"(5x)/(2))/("sin"(x)/(2))dx` is equal to (where, C is a constant of integration)

Options

A. `2x + "sin" x + 2 "sin" 2x + C`
B. `x + 2"sin" x + 2 "sin" 2x + C`
C. `x + 2"sin" x + "sin" 2x + C`
D. `2x + "sin" x + "sin" 2x + C`

Explanation

To solve the integral \( \int \frac{\sin(5x/2)}{\sin(x/2)} \, dx \), we will follow a series of steps involving trigonometric identities and integration techniques.

### Step-by-Step Solution:

1. **Rewrite the Integral**:
\[ I = \int \frac{\sin(5x/2)}{\sin(x/2)} \, dx \]

2. **Multiply by a Convenient Factor**:
Multiply the numerator and denominator by \( 2 \cos(x/2) \):
\[ I = \int \frac{2 \sin(5x/2) \cos(x/2)}{2 \sin(x/2) \cos(x/2)} \, dx \]

3. **Use Trigonometric Identities**:
Using the identity \( 2 \sin A \cos B = \sin(A + B) + \sin(A - B) \):
- For the numerator: \( 2 \sin(5x/2) \cos(x/2) = \sin(5x/2 + x/2) + \sin(5x/2 - x/2) = \sin(3x) + \sin(2x) \)
- For the denominator: \( 2 \sin(x/2) \cos(x/2) = \sin(x) \)

Thus, we can rewrite the integral as:
\[ I = \int \frac{\sin(3x) + \sin(2x)}{\sin(x)} \, dx \]

4. **Split the Integral**:
\[ I = \int \frac{\sin(3x)}{\sin(x)} \, dx + \int \frac{\sin(2x)}{\sin(x)} \, dx \]

5. **Use Known Integrals**:
We can use the known integrals:
- \( \int \frac{\sin(nx)}{\sin(x)} \, dx = n x - \frac{1}{2} \sum_{k=1}^{n-1} \frac{\sin((n-2k)x)}{n-2k} + C \)

For \( n = 3 \):
\[ \int \frac{\sin(3x)}{\sin(x)} \, dx = 3x - \frac{1}{2} \left( \frac{\sin(x)}{1} \right) + C_1 \]

For \( n = 2 \):
\[ \int \frac{\sin(2x)}{\sin(x)} \, dx = 2x - \frac{1}{2} \left( \frac{\sin(0)}{0} \right) + C_2 \]

6. **Combine the Results**:
Combine the results of both integrals:
\[ I = \left(3x - \frac{1}{2} \sin(x)\right) + \left(2x\right) + C \]
Simplifying gives:
\[ I = 5x - \frac{1}{2} \sin(x) + C \]

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