`I=int_(0)^(2pi)(1)/(1+e^(sinx))dx` is equal to

Options

A. `pi`
B. `2pi`
C. `(pi)/(2)`
D. none of these

Explanation

To solve the integral \( I = \int_{0}^{2\pi} \frac{1}{1 + e^{\sin x}} \, dx \), we can use a property of definite integrals.

### Step-by-Step Solution:

1. **Using the Property of Definite Integrals:**
We can use the property that states:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
Here, \( a = 0 \) and \( b = 2\pi \). Therefore, we can rewrite the integral as:
\[ I = \int_{0}^{2\pi} \frac{1}{1 + e^{\sin(2\pi - x)}} \, dx \]

2. **Simplifying the Integral:**
Since \( \sin(2\pi - x) = -\sin x \), we can substitute this into our integral:
\[ I = \int_{0}^{2\pi} \frac{1}{1 + e^{-\sin x}} \, dx \]

3. **Rewriting the Integral:**
We can rewrite \( \frac{1}{1 + e^{-\sin x}} \) as:
\[ \frac{1}{1 + e^{-\sin x}} = \frac{e^{\sin x}}{e^{\sin x} + 1} \]
Thus, we have:
\[ I = \int_{0}^{2\pi} \frac{e^{\sin x}}{e^{\sin x} + 1} \, dx \]

4. **Adding the Two Expressions for \( I \):**
Now we have two expressions for \( I \):
\[ I = \int_{0}^{2\pi} \frac{1}{1 + e^{\sin x}} \, dx \quad \text{(Equation 1)} \]
\[ I = \int_{0}^{2\pi} \frac{e^{\sin x}}{e^{\sin x} + 1} \, dx \quad \text{(Equation 2)} \]
Adding these two equations:
\[ 2I = \int_{0}^{2\pi} \left( \frac{1}{1 + e^{\sin x}} + \frac{e^{\sin x}}{e^{\sin x} + 1} \right) dx \]

5. **Simplifying the Sum:**
The sum inside the integral simplifies to:
\[ \frac{1 + e^{\sin x}}{1 + e^{\sin x}} = 1 \]
Therefore:
\[ 2I = \int_{0}^{2\pi} 1 \, dx \]

6. **Calculating the Integral:**
The integral of 1 from 0 to \( 2\pi \) is simply:
\[ \int_{0}^{2\pi} 1 \, dx = 2\pi \]
Thus:
\[ 2I = 2\pi \]

7. **Finding \( I \):**
Dividing both sides by 2 gives:
\[ I = \pi \]

### Final Result:
The value of the integral \( I \) is:
\[ I = \pi \]