Evaluate Definite Integral with Exponential and Trigonometric Terms

To evaluate the definite integral $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1 + e^x} \, dx$

we apply the property of definite integrals: $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx$

Here, $a = -\frac{\pi}{2}$ and $b = \frac{\pi}{2}$, which gives $a + b - x = -x$. Replacing $x$ with $-x$ in the integral, we get: $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(-x)^2 \cos(-x)}{1 + e^{-x}} \, dx$

Since $(-x)^2 = x^2$ and $\cos(-x) = \cos x$, this simplifies to: $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1 + \frac{1}{e^x}} \, dx$ $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{e^x x^2 \cos x}{e^x + 1} \, dx$

Adding the original integral and this transformed integral together: $2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1 + e^x} \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{e^x x^2 \cos x}{1 + e^x} \, dx$ $2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(1 + e^x) x^2 \cos x}{1 + e^x} \, dx$ $2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \, dx$

Since the function $f(x) = x^2 \cos x$ is an even function ($f(-x) = f(x)$), we can write: $2I = 2 \int_{0}^{\frac{\pi}{2}} x^2 \cos x \, dx$ $I = \int_{0}^{\frac{\pi}{2}} x^2 \cos x \, dx$

Now, using integration by parts: $\int u \, dv = uv - \int v \, du$

Let $u = x^2 \implies du = 2x \, dx$ and $dv = \cos x \, dx \implies v = \sin x$.

$I = \left[ x^2 \sin x \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2x \sin x \, dx$

Evaluating the first term: $\left[ x^2 \sin x \right]_{0}^{\frac{\pi}{2}} = \left( \frac{\pi}{2} \right)^2 \sin\left(\frac{\pi}{2}\right) - 0 = \frac{\pi^2}{4}$

Next, integrate $\int_{0}^{\frac{\pi}{2}} 2x \sin x \, dx$ by parts again with $u = 2x$ and $dv = \sin x \, dx$: $\int_{0}^{\frac{\pi}{2}} 2x \sin x \, dx = \left[ -2x \cos x \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} (-2 \cos x) \, dx$ $= \left[ -2x \cos x + 2 \sin x \right]_{0}^{\frac{\pi}{2}}$ $= \left( -2\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) + 2 \sin\left(\frac{\pi}{2}\right) \right) - \left( 0 + 2 \sin 0 \right)$ $= 0 + 2(1) - 0 = 2$

Substituting this back into the expression for $I$: $I = \frac{\pi^2}{4} - 2$

Thus, the correct option is A.