To solve the integral
\[
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \sin^2 x}{1 + \pi^{\sin x}} \, dx,
\]
we will follow these steps:
### Step 1: Define the integral
Let
\[
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \sin^2 x}{1 + \pi^{\sin x}} \, dx.
\]
### Step 2: Substitute \( x \) with \( -x \)
We will replace \( x \) with \( -x \) in the integral:
\[
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \sin^2(-x)}{1 + \pi^{\sin(-x)}} \, dx.
\]
Using the properties of sine, we know that \( \sin(-x) = -\sin x \). Thus, we have:
\[
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \sin^2 x}{1 + \pi^{-\sin x}} \, dx.
\]
### Step 3: Simplify the denominator
The term \( \pi^{-\sin x} \) can be rewritten as \( \frac{1}{\pi^{\sin x}} \). Therefore, we can express the integral as:
\[
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \sin^2 x}{1 + \frac{1}{\pi^{\sin x}}} \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(1 + \sin^2 x) \cdot \pi^{\sin x}}{\pi^{\sin x} + 1} \, dx.
\]
### Step 4: Combine the two expressions for \( I \)
Now we have two expressions for \( I \):
1. \( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \sin^2 x}{1 + \pi^{\sin x}} \, dx \)
2. \( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(1 + \sin^2 x) \cdot \pi^{\sin x}}{\pi^{\sin x} + 1} \, dx \)
### Step 5: Add the two expressions
Adding these two equations gives:
\[
2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{1 + \sin^2 x}{1 + \pi^{\sin x}} + \frac{(1 + \sin^2 x) \cdot \pi^{\sin x}}{\pi^{\sin x} + 1} \right) dx.
\]
### Step 6: Simplify the integrand
The integrand simplifies to:
\[
\frac{(1 + \sin^2 x)(1 + \pi^{\sin x})}{1 + \pi^{\sin x}} = 1 + \sin^2 x.
\]
Thus, we have:
\[
2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1 + \sin^2 x) \, dx.
\]
### Step 7: Evaluate the integral
Now we can split the integral:
\[
2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx.
\]
The first integral evaluates to:
\[
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx = \pi.
\]
For the second integral, we use the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \):
\[
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \left( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(2x) \, dx \right).
\]
The integral of \( \cos(2x) \) over a symmetric interval around zero is zero, thus:
\[
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{1}{2} \cdot \pi = \frac{\pi}{2}.
\]
### Step 8: Combine results
Now substituting back, we have:
\[
2I = \pi + \frac{\pi}{2} = \frac{3\pi}{2}.
\]
Thus, dividing by 2 gives:
\[
I = \frac{3\pi}{4}.
\]
### Final Answer
The value of the integral is
\[
\boxed{\frac{3\pi}{4}}.
\]
