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

Options

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

Explanation

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 1: Understand the integral

The integral \( a_n \) involves the function \( \sin^2((n+1)x) \) divided by \( \sin(2x) \). We need to analyze how \( a_n \) behaves as \( n \) changes.

### Step 2: Recognize the pattern

From the video transcript, it is indicated that the sequence \( a_n \) behaves like an arithmetic progression (AP) with a common difference of 0. This suggests that the values of \( a_n \) do not change with \( n \).

### Step 3: Establish the equality

Since \( a_n \) is constant for all \( n \), we can express this as:

\[ a_n = a_{n-1} = a_{n-2} = a_{n-3} = a_{n-4} = C, \]

where \( C \) is some constant value.

### Step 4: Calculate \( a_{n-1} - a_{n-4} \)

Now, substituting into the expression we want to evaluate:

\[ a_{n-1} - a_{n-4} = C - C = 0. \]

### Conclusion

Thus, we find that:

\[ a_{n-1} - a_{n-4} = 0. \]