For any `n in N, int_(0)^(pi) (sin (2n+1)x)/(sinx)dx` is equal to

Options

A. `pi`
B. 0
C. `n pi`
D. `(2n+1)=pi`

Explanation

To solve the integral \( I_n = \int_0^{\pi} \frac{\sin(2n+1)x}{\sin x} \, dx \), we will use a recursive approach.

### Step 1: Define the Integral
Let
\[ I_n = \int_0^{\pi} \frac{\sin(2n+1)x}{\sin x} \, dx \]

### Step 2: Write the Next Integral
We can express \( I_{n+1} \) as follows:
\[ I_{n+1} = \int_0^{\pi} \frac{\sin(2(n+1)+1)x}{\sin x} \, dx = \int_0^{\pi} \frac{\sin(2n+3)x}{\sin x} \, dx \]

### Step 3: Subtract the Two Integrals
Now, we will subtract \( I_n \) from \( I_{n+1} \):
\[ I_{n+1} - I_n = \int_0^{\pi} \left( \frac{\sin(2n+3)x - \sin(2n+1)x}{\sin x} \right) \, dx \]

### Step 4: Use the Sine Difference Formula
Using the sine difference formula:
\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
we have:
\[ \sin(2n+3)x - \sin(2n+1)x = 2 \cos\left((2n+3 + 2n+1)x/2\right) \sin\left((2n+3 - (2n+1))x/2\right) \]
This simplifies to:
\[ = 2 \cos((4n+4)x/2) \sin(x) \]
Thus,
\[ I_{n+1} - I_n = \int_0^{\pi} \frac{2 \cos((2n+2)x) \sin x}{\sin x} \, dx = 2 \int_0^{\pi} \cos((2n+2)x) \, dx \]

### Step 5: Evaluate the Integral
The integral \( \int_0^{\pi} \cos((2n+2)x) \, dx \) evaluates to:
- If \( 2n+2 \neq 0 \), it equals \( 0 \).
- If \( 2n+2 = 0 \), which is not possible for \( n \in \mathbb{N} \).

Thus, we have:
\[ I_{n+1} - I_n = 0 \quad \Rightarrow \quad I_{n+1} = I_n \]

### Step 6: Base Case
Now we can find a base case. Let’s calculate \( I_1 \):
\[ I_1 = \int_0^{\pi} \frac{\sin(3x)}{\sin x} \, dx \]
Using the identity \( \sin(3x) = 3\sin x - 4\sin^3 x \):
\[ I_1 = \int_0^{\pi} \frac{3\sin x - 4\sin^3 x}{\sin x} \, dx = \int_0^{\pi} (3 - 4\sin^2 x) \, dx \]

### Step 7: Evaluate \( I_1 \)
Now, we can evaluate:
\[ I_1 = \int_0^{\pi} 3 \, dx - 4 \int_0^{\pi} \sin^2 x \, dx \]
The first integral evaluates to:
\[ 3 \cdot \pi \]
The second integral can be computed using:
\[ \int_0^{\pi} \sin^2 x \, dx = \frac{\pi}{2} \]
Thus:
\[ I_1 = 3\pi - 4 \cdot \frac{\pi}{2} = 3\pi - 2\pi = \pi \]

### Conclusion
Since \( I_n = I_1 \) for all \( n \in \mathbb{N} \), we conclude:
\[ I_n = \pi \]

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