The value of ` I(n) =int_0^pi(sin ^2n theta)/(sin^2theta) d theta ` is `(AA n in N)`

Options

A. (a)`npi`
B. (b)`(npi)/(2)`
C. (c)`(npi)/(4)`
D. (d)None of these

Answer

To solve the integral \( I(n) = \int_0^\pi \frac{\sin^2(n \theta)}{\sin^2(\theta)} d\theta \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Substituting the Integral**:
We start with the integral:
\[ I(n) = \int_0^\pi \frac{\sin^2(n \theta)}{\sin^2(\theta)} d\theta \]

2. **Evaluating for \( n = 1 \)**:
Let's first evaluate \( I(1) \):
\[ I(1) = \int_0^\pi \frac{\sin^2(\theta)}{\sin^2(\theta)} d\theta = \int_0^\pi 1 \, d\theta \]
This simplifies to:
\[ I(1) = \left[ \theta \right]_0^\pi = \pi \]

3. **Evaluating for \( n = 2 \)**:
Next, we evaluate \( I(2) \):
\[ I(2) = \int_0^\pi \frac{\sin^2(2\theta)}{\sin^2(\theta)} d\theta \]
We can use the identity \( \sin^2(2\theta) = 4\sin^2(\theta)\cos^2(\theta) \):
\[ I(2) = \int_0^\pi \frac{4\sin^2(\theta)\cos^2(\theta)}{\sin^2(\theta)} d\theta = 4 \int_0^\pi \cos^2(\theta) d\theta \]

4. **Integrating \( \cos^2(\theta) \)**:
We can use the identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \):
\[ I(2) = 4 \int_0^\pi \frac{1 + \cos(2\theta)}{2} d\theta = 2 \int_0^\pi (1 + \cos(2\theta)) d\theta \]
Evaluating this gives:
\[ = 2 \left[ \theta + \frac{\sin(2\theta)}{2} \right]_0^\pi = 2 \left[ \pi + 0 \right] = 2\pi \]

5. **Generalizing for \( n \)**:
From our evaluations, we observe that:
\[ I(1) = \pi, \quad I(2) = 2\pi \]
It appears that \( I(n) = n\pi \) for natural numbers \( n \).

6. **Conclusion**:
Therefore, we conclude that:
\[ I(n) = n\pi \quad \text{for } n \in \mathbb{N} \]

### Final Answer:
\[ I(n) = n\pi \]