1. Let $I=\int_{-\pi/2}^{\pi/2} \frac{1}{1+e^{\sin x}}\,dx.$

2. Use the substitution $x\mapsto -x$. Since the limits $-\pi/2$ to $\pi/2$ are symmetric,

   =\int_{-\pi/2}^{\pi/2} \frac{1}{1+e^{-\sin x}}\,dx.$$

3. Now add the two expressions for $I$: $2I=\int_{-\pi/2}^{\pi/2} \left(\frac{1}{1+e^{\sin x}}+\frac{1}{1+e^{-\sin x}}\right)dx.$

4. Simplify the bracket. Let $a=e^{\sin x}$. Then

   =\frac{1}{1+a}+\frac{a}{1+a}=1.$$ Hence, $$\frac{1}{1+e^{\sin x}}+\frac{1}{1+e^{-\sin x}}=1.$$

5. Therefore,

   =\left[x\right]_{-\pi/2}^{\pi/2} =\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi.$$

6. So, $I=\frac{\pi}{2}.$

7. Checking options:

   • A: $\pi$ ❌
   • B: $\frac{3\pi}{2}$ ❌
   • C: $\frac{\pi}{2}$ ✅
   • D: $\frac{\pi}{4}$ ❌

Thus the correct answer is Option C.