If `I=int_(-pi/2)^(pi/2) 1/(1+e^(sinx))dx` then `I` is

Answer

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

### Step-by-Step Solution:

1. **Use the property of definite integrals**:
We know that:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
For our case, \( a = -\frac{\pi}{2} \) and \( b = \frac{\pi}{2} \). Thus, we can write:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{1 + e^{\sin x}} \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{1 + e^{\sin(-x)}} \, dx \]

2. **Simplify the expression**:
Since \( \sin(-x) = -\sin x \), we have:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{1 + e^{-\sin x}} \, dx \]

3. **Rewrite the integral**:
We can rewrite the integral:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{1 + \frac{1}{e^{\sin x}}} \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{e^{\sin x}}{e^{\sin x} + 1} \, dx \]

4. **Combine the two expressions for \( I \)**:
Now we have two expressions for \( I \):
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{1 + e^{\sin x}} \, dx \]
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{e^{\sin x}}{e^{\sin x} + 1} \, dx \]

5. **Add the two integrals**:
Adding these two integrals gives:
\[ 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{1}{1 + e^{\sin x}} + \frac{e^{\sin x}}{e^{\sin x} + 1} \right) \, dx \]
Simplifying the integrand:
\[ \frac{1 + e^{\sin x}}{1 + e^{\sin x}} = 1 \]
Therefore:
\[ 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx \]

6. **Evaluate the integral**:
The integral of 1 from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) is:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi \]
Thus:
\[ 2I = \pi \implies I = \frac{\pi}{2} \]

### Final Answer:
\[ I = \frac{\pi}{2} \]