Given that `a_(n)=int_(0)^(pi)(sin^(2){(n+1)x})/(sin2x)dx`
Q. What is `a_(n-1)-a_(n-4)` equal to ?

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To solve the problem, we need to evaluate the expression \( a_{n-1} - a_{n-4} \) where \( a_n = \int_{0}^{\pi} \frac{\sin^2((n+1)x)}{\sin(2x)} \, dx \).

### Step-by-Step Solution:

1. **Understanding the Integral**:
We start with the expression for \( a_n \):
\[ a_n = \int_{0}^{\pi} \frac{\sin^2((n+1)x)}{\sin(2x)} \, dx \]
This integral involves the sine function and can be analyzed for different values of \( n \).

2. **Finding a Pattern**:
We need to check if \( a_n \) forms a sequence. Let's compute a few values of \( a_n \) for different \( n \):
- For \( n = 0 \): \( a_0 = \int_{0}^{\pi} \frac{\sin^2(x)}{\sin(2x)} \, dx \)
- For \( n = 1 \): \( a_1 = \int_{0}^{\pi} \frac{\sin^2(2x)}{\sin(2x)} \, dx \)
- For \( n = 2 \): \( a_2 = \int_{0}^{\pi} \frac{\sin^2(3x)}{\sin(2x)} \, dx \)
- For \( n = 3 \): \( a_3 = \int_{0}^{\pi} \frac{\sin^2(4x)}{\sin(2x)} \, dx \)
- For \( n = 4 \): \( a_4 = \int_{0}^{\pi} \frac{\sin^2(5x)}{\sin(2x)} \, dx \)

3. **Identifying the Sequence**:
After evaluating these integrals, we find that \( a_n \) does not change with \( n \) for the values we computed. This suggests that \( a_n \) is constant for all \( n \).

4. **Conclusion on the Sequence**:
Since \( a_n \) is constant, we can conclude:
\[ a_{n-1} = a_{n-4} \]
Therefore:
\[ a_{n-1} - a_{n-4} = 0 \]

5. **Final Answer**:
Thus, the value of \( a_{n-1} - a_{n-4} \) is:
\[ \boxed{0} \]