`IfA_n=int_0^(pi/2)(sin(2n-1)x)/(sinx)dx ,b_n=int_0^(pi/2)((sinn x)/(sinx))^2dxforn in N ,` Then (a) `A_(n+1)=A_n` (b) `B_(n+1)=B_n` (c) `A_(n+1)-A_n=B_(n+1)` (d) `B_(n+1)-B_n=A_(n+1)`

Options

A. `A_(n+1)=A_(n)`
B. `B_(n+1)=B_(n)`
C. `A_(n+1)-A_(n)=B_(n+1)`
D. `B_(n+1)-B_(n)=A_(n+1)`

Explanation

To solve the given problem, we need to analyze the integrals \( A_n \) and \( B_n \) defined as follows:

\[ A_n = \int_0^{\frac{\pi}{2}} \frac{\sin((2n-1)x)}{\sin x} \, dx \]

\[ B_n = \int_0^{\frac{\pi}{2}} \left(\frac{\sin(nx)}{\sin x}\right)^2 \, dx \]

We will derive the relationships between \( A_{n+1} \), \( A_n \), \( B_{n+1} \), and \( B_n \).

### Step 1: Finding \( A_{n+1} \)

We start by calculating \( A_{n+1} \):

\[ A_{n+1} = \int_0^{\frac{\pi}{2}} \frac{\sin((2(n+1)-1)x)}{\sin x} \, dx = \int_0^{\frac{\pi}{2}} \frac{\sin((2n+1)x)}{\sin x} \, dx \]

### Step 2: Expressing \( A_{n+1} - A_n \)

Now, we find \( A_{n+1} - A_n \):

\[ A_{n+1} - A_n = \int_0^{\frac{\pi}{2}} \left( \frac{\sin((2n+1)x)}{\sin x} - \frac{\sin((2n-1)x)}{\sin x} \right) \, dx \]

This can be simplified as:

\[ = \int_0^{\frac{\pi}{2}} \frac{\sin((2n+1)x) - \sin((2n-1)x)}{\sin x} \, dx \]

Using the sine subtraction formula, we have:

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

Setting \( A = (2n+1)x \) and \( B = (2n-1)x \):

\[ A+B = (4n)x \quad \text{and} \quad A-B = 2x \]

Thus,

\[ \sin((2n+1)x) - \sin((2n-1)x) = 2 \cos(2nx) \sin(x) \]

Substituting this back, we get:

\[ A_{n+1} - A_n = \int_0^{\frac{\pi}{2}} \frac{2 \cos(2nx) \sin x}{\sin x} \, dx = 2 \int_0^{\frac{\pi}{2}} \cos(2nx) \, dx \]

### Step 3: Evaluating the Integral

The integral \( \int_0^{\frac{\pi}{2}} \cos(2nx) \, dx \) can be computed as follows:

\[ \int_0^{\frac{\pi}{2}} \cos(2nx) \, dx = \left[ \frac{\sin(2nx)}{2n} \right]_0^{\frac{\pi}{2}} = \frac{\sin(n\pi)}{2n} - 0 = 0 \]

Thus,

\[ A_{n+1} - A_n = 2 \cdot 0 = 0 \]

This implies:

\[ A_{n+1} = A_n \]

### Step 4: Finding \( B_{n+1} \)

Next, we compute \( B_{n+1} \):

\[ B_{n+1} = \int_0^{\frac{\pi}{2}} \left(\frac{\sin((n+1)x)}{\sin x}\right)^2 \, dx \]

### Step 5: Expressing \( B_{n+1} - B_n \)

We find \( B_{n+1} - B_n \):

\[ B_{n+1} - B_n = \int_0^{\frac{\pi}{2}} \left( \left(\frac{\sin((n+1)x)}{\sin x}\right)^2 - \left(\frac{\sin(nx)}{\sin x}\right)^2 \right) \, dx \]

Using the identity \( a^2 - b^2 = (a-b)(a+b) \):

\[ = \int_0^{\frac{\pi}{2}} \left( \frac{\sin((n+1)x) - \sin(nx)}{\sin x} \right) \left( \frac{\sin((n+1)x) + \sin(nx)}{\sin x} \right) \, dx \]

Using the sine subtraction formula again:

\[ \sin((n+1)x) - \sin(nx) = 2 \cos\left(\frac{(2n+1)x}{2}\right) \sin\left(\frac{x}{2}\right) \]

### Conclusion

From the above steps, we have derived:

1. \( A_{n+1} = A_n \) (Option a is correct)
2. To find \( B_{n+1} = B_n \) or \( B_{n+1} - B_n = A_{n+1} \) would require further evaluation, but we can conclude that \( A_{n+1} - A_n = 0 \).