For any `n in N`, the value of the intergral `int_(0)^(pi) (sin 2nx)/(sinx)dx` is,

Options

A. `pi`
B. `2pi`
C. `-pi`
D. none of these

Explanation

To solve the integral \( I = \int_{0}^{\pi} \frac{\sin(2nx)}{\sin(x)} \, dx \) for any \( n \in \mathbb{N} \), we will evaluate it step by step.

### Step 1: Define the Integral
Let
\[ I = \int_{0}^{\pi} \frac{\sin(2nx)}{\sin(x)} \, dx \]

### Step 2: Evaluate for \( n = 1 \)
For \( n = 1 \):
\[ I_1 = \int_{0}^{\pi} \frac{\sin(2x)}{\sin(x)} \, dx \]
Using the identity \( \sin(2x) = 2 \sin(x) \cos(x) \):
\[ I_1 = \int_{0}^{\pi} \frac{2 \sin(x) \cos(x)}{\sin(x)} \, dx = 2 \int_{0}^{\pi} \cos(x) \, dx \]
The integral of \( \cos(x) \) from \( 0 \) to \( \pi \) is:
\[ \int_{0}^{\pi} \cos(x) \, dx = [\sin(x)]_{0}^{\pi} = \sin(\pi) - \sin(0) = 0 \]
Thus,
\[ I_1 = 2 \cdot 0 = 0 \]

### Step 3: Evaluate for \( n = 2 \)
For \( n = 2 \):
\[ I_2 = \int_{0}^{\pi} \frac{\sin(4x)}{\sin(x)} \, dx \]
Using the identity \( \sin(4x) = 2 \sin(2x) \cos(2x) \):
\[ I_2 = \int_{0}^{\pi} \frac{2 \sin(2x) \cos(2x)}{\sin(x)} \, dx \]
Using \( \sin(2x) = 2 \sin(x) \cos(x) \):
\[ I_2 = 2 \int_{0}^{\pi} \frac{2 \sin(x) \cos(x) \cos(2x)}{\sin(x)} \, dx = 4 \int_{0}^{\pi} \cos(x) \cos(2x) \, dx \]
Using the product-to-sum identities:
\[ \cos(x) \cos(2x) = \frac{1}{2} [\cos(x) + \cos(3x)] \]
Thus,
\[ I_2 = 4 \cdot \frac{1}{2} \left( \int_{0}^{\pi} \cos(x) \, dx + \int_{0}^{\pi} \cos(3x) \, dx \right) \]
Both integrals evaluate to zero:
\[ \int_{0}^{\pi} \cos(x) \, dx = 0, \quad \int_{0}^{\pi} \cos(3x) \, dx = 0 \]
Thus,
\[ I_2 = 0 \]

### Step 4: General Case for \( n \)
By observing the pattern, we see that for any \( n \in \mathbb{N} \):
\[ I_n = 0 \]
This is because each integral \( I_n \) evaluates to zero due to the periodic nature of the sine and cosine functions over the interval \( [0, \pi] \).

### Conclusion
Therefore, the value of the integral
\[ \int_{0}^{\pi} \frac{\sin(2nx)}{\sin(x)} \, dx = 0 \quad \text{for any } n \in \mathbb{N}. \]