The value of $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\frac{1 + \sin^2 x}{1 + \pi^{\sin x}}\right) dx$$ is:

Let us denote the required integral by

$$I \;=\;\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{1+\sin^2x}{\,1+\pi^{\sin x}\,}\,dx.$$

We put

$$f(x)=\dfrac{1+\sin^2x}{1+\pi^{\sin x}}.$$

Because the limits are symmetric about the origin, we use the standard property

$$$\int_{-a}^{a}f(x)\,dx=\frac12\int_{-a}^{a}\bigl[f(x)+f(-x)\bigr]\,dx.$$$

So we first find $$f(-x):$$

Since $$\sin(-x)=-\sin x$$ and $$\sin^2(-x)=\sin^2x,$$ we have

$$f(-x)=\dfrac{1+\sin^2x}{1+\pi^{\sin(-x)}}=\dfrac{1+\sin^2x}{1+\pi^{-\sin x}}.$$

Now we add $$f(x)$$ and $$f(-x):$$

$$$\begin{aligned} f(x)+f(-x) &=\dfrac{1+\sin^2x}{1+\pi^{\sin x}}+\dfrac{1+\sin^2x}{1+\pi^{-\sin x}}\\[4pt] &=(1+\sin^2x)\!\left[\dfrac1{1+\pi^{\sin x}}+\dfrac1{1+\pi^{-\sin x}}\right]. \end{aligned}$$$

Write $$t=\pi^{\sin x} \;(\;t>0\;).$$ Then $$\pi^{-\sin x}=t^{-1}.$$ Inside the bracket we get

$$$\dfrac1{1+t}+\dfrac1{1+t^{-1}} =\dfrac1{1+t}+\dfrac t{t+1} =\dfrac{1+t}{1+t}=1.$$$

Hence

$$f(x)+f(-x)=1+\sin^2x.$$

Using the property of symmetrical limits, we can now write

$$$I=\frac12\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\bigl[f(x)+f(-x)\bigr]\,dx =\frac12\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\bigl(1+\sin^2x\bigr)\,dx.$$$

We split the integral:

$$I=\frac12\left[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1\,dx +\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2x\,dx\right].$$

The first integral is simply the length of the interval:

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1\,dx=\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)=\pi.$$

For the second integral, we use the identity $$\sin^2x=\dfrac{1-\cos2x}{2}.$$ Stating this formula explicitly allows us to integrate easily:

$$$\begin{aligned} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2x\,dx &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{1-\cos2x}{2}\,dx\\[4pt] &=\frac12\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dx-\frac12\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos2x\,dx. \end{aligned}$$$

The first part gives $$\frac12\cdot\pi=\frac{\pi}{2}.$$ For the second part, note that $$\int\cos2x\,dx=\frac{\sin2x}{2},$$ and $$\sin2x$$ is zero at both $$x=\frac{\pi}{2}$$ and $$x=-\frac{\pi}{2},$$ so the term vanishes. Therefore,

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

Substituting these results back, we obtain

$$I=\frac12\left[\pi+\frac{\pi}{2}\right]=\frac12\left(\frac{3\pi}{2}\right)=\frac{3\pi}{4}.$$

Hence, the correct answer is Option D.