The value of the integral `int_(0)^(npi+v)|sinx|dx" where "ninNand0levlepi` is

Options

A. `(2n+1-sinv)`
B. `(2n-1-cosv)`
C. `(2n+1-cosv)`
D. none of these

Answer

To solve the integral \( I = \int_{0}^{n\pi + v} |\sin x| \, dx \), where \( n \in \mathbb{N} \) and \( 0 \leq v \leq \pi \), we will break down the integral into manageable parts.

### Step-by-Step Solution:

1. **Break the Integral into Parts**:
Since \( |\sin x| \) is periodic with a period of \( 2\pi \), we can break the integral into two parts:
\[ I = \int_{0}^{v} |\sin x| \, dx + \int_{v}^{n\pi + v} |\sin x| \, dx \]

2. **Evaluate the First Integral**:
For the first integral from \( 0 \) to \( v \), since \( v \) is between \( 0 \) and \( \pi \), \( |\sin x| = \sin x \) in this interval:
\[ \int_{0}^{v} |\sin x| \, dx = \int_{0}^{v} \sin x \, dx \]
The integral of \( \sin x \) is:
\[ -\cos x \bigg|_{0}^{v} = -\cos v + \cos 0 = 1 - \cos v \]

3. **Evaluate the Second Integral**:
For the second integral from \( v \) to \( n\pi + v \), we need to determine how many complete periods of \( 2\pi \) fit into this range. The integral can be expressed as:
\[ \int_{v}^{n\pi + v} |\sin x| \, dx = \int_{v}^{2\pi + v} |\sin x| \, dx + \int_{2\pi + v}^{n\pi + v} |\sin x| \, dx \]
The integral \( \int_{v}^{2\pi + v} |\sin x| \, dx \) can be computed as follows:
- From \( v \) to \( 2\pi \), \( |\sin x| = -\sin x \) (since \( \sin x \) is negative in this range).
- From \( 2\pi \) to \( 2\pi + v \), \( |\sin x| = \sin x \).

Thus, we have:
\[ \int_{v}^{2\pi + v} |\sin x| \, dx = -\int_{v}^{2\pi} \sin x \, dx + \int_{2\pi}^{2\pi + v} \sin x \, dx \]

4. **Calculate the Integral Over One Period**:
The integral over one complete period \( [0, 2\pi] \) is:
\[ \int_{0}^{2\pi} |\sin x| \, dx = 2 \int_{0}^{\pi} \sin x \, dx = 2 \cdot (1 - 0) = 2 \]

5. **Combine the Results**:
The integral from \( 0 \) to \( n\pi + v \) can be computed as:
\[ I = (n \text{ complete periods}) \cdot 2 + (1 - \cos v) + \text{(additional parts)} \]
The additional parts will depend on the specific value of \( v \) and how it fits into the periodic structure.

6. **Final Expression**:
After evaluating all parts, we can express the final value of the integral as:
\[ I = 2n + 1 - \cos v \]

### Final Answer:
The value of the integral \( \int_{0}^{n\pi + v} |\sin x| \, dx \) is:
\[ I = 2n + 1 - \cos v \]