If `I_(n) = int_(0)^(pi//2) (sin^(2)nx)/(sin^(2)x)dx` then `I_(1),I_(2),I_(3),...` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Answer

To solve the problem, we need to analyze the integral \( I_n = \int_0^{\frac{\pi}{2}} \frac{\sin^2(nx)}{\sin^2(x)} \, dx \) and show that \( I_1, I_2, I_3, \ldots \) are in arithmetic progression (AP).

### Step-by-Step Solution:

1. **Define the Integral**:
We start with the integral:
\[ I_n = \int_0^{\frac{\pi}{2}} \frac{\sin^2(nx)}{\sin^2(x)} \, dx \]

2. **Express the Difference**:
We compute the difference between consecutive terms:
\[ I_n - I_{n-1} = \int_0^{\frac{\pi}{2}} \left( \frac{\sin^2(nx)}{\sin^2(x)} - \frac{\sin^2((n-1)x)}{\sin^2(x)} \right) \, dx \]
This simplifies to:
\[ I_n - I_{n-1} = \int_0^{\frac{\pi}{2}} \frac{\sin^2(nx) - \sin^2((n-1)x)}{\sin^2(x)} \, dx \]

3. **Use the Identity for Sine**:
We can use the identity \( \sin^2 A - \sin^2 B = (\sin A - \sin B)(\sin A + \sin B) \):
\[ \sin^2(nx) - \sin^2((n-1)x) = (\sin(nx) - \sin((n-1)x))(\sin(nx) + \sin((n-1)x)) \]

4. **Simplify the Integral**:
The integral can be expressed using the sine difference:
\[ I_n - I_{n-1} = \int_0^{\frac{\pi}{2}} \frac{\sin(nx) - \sin((n-1)x)}{\sin(x)} \cdot (\sin(nx) + \sin((n-1)x)) \, dx \]

5. **Evaluate the Integral**:
The integral \( \int_0^{\frac{\pi}{2}} \frac{\sin(nx) - \sin((n-1)x)}{\sin(x)} \, dx \) can be evaluated using known results, leading to:
\[ I_n - I_{n-1} = \frac{\pi}{2} \]

6. **Conclude the Arithmetic Progression**:
Since \( I_n - I_{n-1} = \frac{\pi}{2} \), we can conclude that:
\[ I_2 - I_1 = I_3 - I_2 = I_4 - I_3 = \ldots = \frac{\pi}{2} \]
This means that \( I_1, I_2, I_3, \ldots \) are in arithmetic progression.

### Final Answer:
Thus, \( I_1, I_2, I_3, \ldots \) are in arithmetic progression.

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