If $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{96x^{2}\cos^{2}x}{(1+e^{x})}dx=\pi (\alpha \pi^{2}+\beta),\alpha,\beta \in \mathbb{Z}$$,then $$(\alpha +\beta)^{2}$$ equals

The given integral is $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{96x^{2}\cos^{2}x}{(1+e^{x})}dx$$.

Define the function $$f(x) = \frac{96x^{2}\cos^{2}x}{1+e^{x}}$$.

Since the limits are symmetric, use the property: $$\int_{-a}^{a} f(x) dx = \int_{0}^{a} [f(x) + f(-x)] dx$$.

Compute $$f(-x)$$:

$$f(-x) = \frac{96(-x)^{2}\cos^{2}(-x)}{1+e^{-x}} = \frac{96x^{2}\cos^{2}x}{1+e^{-x}}$$, as cosine is even.

Note that $$1 + e^{-x} = \frac{e^x + 1}{e^x}$$, so $$\frac{1}{1+e^{-x}} = \frac{e^x}{1+e^x}$$.

Thus, $$f(-x) = 96x^{2}\cos^{2}x \cdot \frac{e^x}{1+e^x} = \frac{96x^{2}\cos^{2}x \cdot e^x}{1+e^x}$$.

Now, compute $$f(x) + f(-x)$$:

$$f(x) + f(-x) = \frac{96x^{2}\cos^{2}x}{1+e^{x}} + \frac{96x^{2}\cos^{2}x \cdot e^x}{1+e^x} = 96x^{2}\cos^{2}x \left( \frac{1}{1+e^x} + \frac{e^x}{1+e^x} \right) = 96x^{2}\cos^{2}x \left( \frac{1 + e^x}{1+e^x} \right) = 96x^{2}\cos^{2}x$$.

Therefore, the integral simplifies to:

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) dx = \int_{0}^{\frac{\pi}{2}} [f(x) + f(-x)] dx = \int_{0}^{\frac{\pi}{2}} 96x^{2}\cos^{2}x dx = 96 \int_{0}^{\frac{\pi}{2}} x^{2} \cos^{2} x dx$$.

Now, evaluate $$\int_{0}^{\frac{\pi}{2}} x^{2} \cos^{2} x dx$$.

Use the identity $$\cos^{2}x = \frac{1 + \cos 2x}{2}$$:

$$\int_{0}^{\frac{\pi}{2}} x^{2} \cos^{2} x dx = \int_{0}^{\frac{\pi}{2}} x^{2} \cdot \frac{1 + \cos 2x}{2} dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} x^{2} dx + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} x^{2} \cos 2x dx$$.

Compute the first integral:

$$\frac{1}{2} \int_{0}^{\frac{\pi}{2}} x^{2} dx = \frac{1}{2} \cdot \frac{x^{3}}{3} \Big|_{0}^{\frac{\pi}{2}} = \frac{1}{6} \left( \frac{\pi}{2} \right)^{3} = \frac{1}{6} \cdot \frac{\pi^{3}}{8} = \frac{\pi^{3}}{48}$$.

Compute the second integral using integration by parts:

Let $$u = x^{2}$$, $$dv = \cos 2x dx$$.

Then $$du = 2x dx$$, $$v = \frac{1}{2} \sin 2x$$.

So, $$\int x^{2} \cos 2x dx = x^{2} \cdot \frac{1}{2} \sin 2x - \int \frac{1}{2} \sin 2x \cdot 2x dx = \frac{x^{2}}{2} \sin 2x - \int x \sin 2x dx$$.

Now, for $$\int x \sin 2x dx$$, use integration by parts again:

Let $$u = x$$, $$dv = \sin 2x dx$$.

Then $$du = dx$$, $$v = -\frac{1}{2} \cos 2x$$.

So, $$\int x \sin 2x dx = x \cdot \left(-\frac{1}{2} \cos 2x\right) - \int \left(-\frac{1}{2} \cos 2x\right) dx = -\frac{x}{2} \cos 2x + \frac{1}{2} \int \cos 2x dx = -\frac{x}{2} \cos 2x + \frac{1}{2} \cdot \frac{1}{2} \sin 2x = -\frac{x}{2} \cos 2x + \frac{1}{4} \sin 2x$$.

Substitute back:

$$\int x^{2} \cos 2x dx = \frac{x^{2}}{2} \sin 2x - \left( -\frac{x}{2} \cos 2x + \frac{1}{4} \sin 2x \right) = \frac{x^{2}}{2} \sin 2x + \frac{x}{2} \cos 2x - \frac{1}{4} \sin 2x$$.

Evaluate from 0 to $$\frac{\pi}{2}$$:

At $$x = \frac{\pi}{2}$$: $$\sin \pi = 0$$, $$\cos \pi = -1$$, so $$\frac{(\pi/2)^{2}}{2} \cdot 0 + \frac{\pi/2}{2} \cdot (-1) - \frac{1}{4} \cdot 0 = -\frac{\pi}{4}$$.

At $$x = 0$$: $$\sin 0 = 0$$, $$\cos 0 = 1$$, so $$\frac{0}{2} \cdot 0 + \frac{0}{2} \cdot 1 - \frac{1}{4} \cdot 0 = 0$$.

Thus, $$\int_{0}^{\frac{\pi}{2}} x^{2} \cos 2x dx = -\frac{\pi}{4} - 0 = -\frac{\pi}{4}$$.

So, the second part is $$\frac{1}{2} \times \left(-\frac{\pi}{4}\right) = -\frac{\pi}{8}$$.

Combine both parts:

$$\int_{0}^{\frac{\pi}{2}} x^{2} \cos^{2} x dx = \frac{\pi^{3}}{48} - \frac{\pi}{8} = \frac{\pi^{3}}{48} - \frac{6\pi}{48} = \frac{\pi^{3} - 6\pi}{48}$$.

Now, multiply by 96:

$$96 \times \frac{\pi^{3} - 6\pi}{48} = 2 (\pi^{3} - 6\pi) = 2\pi^{3} - 12\pi$$.

Thus, the integral equals $$2\pi^{3} - 12\pi$$.

The problem states that this equals $$\pi (\alpha \pi^{2} + \beta)$$, so:

$$2\pi^{3} - 12\pi = \pi (2\pi^{2} - 12)$$.

Therefore, $$\alpha \pi^{2} + \beta = 2\pi^{2} - 12$$, which implies $$\alpha = 2$$ and $$\beta = -12$$.

Now, $$\alpha + \beta = 2 + (-12) = -10$$, and $$(\alpha + \beta)^{2} = (-10)^{2} = 100$$.

The correct option is D. 100.