If the value of the integral $$\int_{-1}^{1} \frac{\cos \alpha x}{1 + 3^x} dx$$ is $$\frac{2}{\pi}$$. Then, a value of $$\alpha$$ is

We need to evaluate $$I = \int_{-1}^{1} \frac{\cos \alpha x}{1 + 3^x} \, dx$$.

Using the property: For a function of the form $$\int_{-a}^{a} \frac{f(x)}{1 + b^x} \, dx$$ where $$f(x)$$ is even, we can use the substitution $$x \to -x$$.

Let $$I = \int_{-1}^{1} \frac{\cos \alpha x}{1 + 3^x} \, dx$$ $$-(1)$$

Substituting $$x \to -x$$: $$I = \int_{-1}^{1} \frac{\cos(-\alpha x)}{1 + 3^{-x}} \, dx = \int_{-1}^{1} \frac{\cos \alpha x}{1 + 3^{-x}} \, dx$$ $$-(2)$$

(since $$\cos$$ is an even function).

Adding $$(1)$$ and $$(2)$$:

$$2I = \int_{-1}^{1} \cos \alpha x \left(\frac{1}{1 + 3^x} + \frac{1}{1 + 3^{-x}}\right) dx$$

Now simplify the expression in parentheses:

$$\frac{1}{1 + 3^x} + \frac{1}{1 + 3^{-x}} = \frac{1}{1 + 3^x} + \frac{3^x}{3^x + 1} = \frac{1 + 3^x}{1 + 3^x} = 1$$

Therefore: $$2I = \int_{-1}^{1} \cos \alpha x \, dx$$

$$2I = \left[\frac{\sin \alpha x}{\alpha}\right]_{-1}^{1} = \frac{\sin \alpha - \sin(-\alpha)}{\alpha} = \frac{2 \sin \alpha}{\alpha}$$

$$I = \frac{\sin \alpha}{\alpha}$$

Given $$I = \frac{2}{\pi}$$:

$$\frac{\sin \alpha}{\alpha} = \frac{2}{\pi}$$

Therefore $$\alpha = \frac{\pi}{2}$$, which matches Option D.