$$\int \frac{\sin^8 x - \cos^8 x}{1 - 2\sin^2 x \cos^2 x} dx$$ is equal to:

We need to solve the integral: $$\int \frac{\sin^8 x - \cos^8 x}{1 - 2\sin^2 x \cos^2 x} dx$$

First, let's simplify the numerator $$\sin^8 x - \cos^8 x$$. We can factor this using the difference of squares formula. Set $$a = \sin^4 x$$ and $$b = \cos^4 x$$, so:

$$\sin^8 x - \cos^8 x = (\sin^4 x)^2 - (\cos^4 x)^2 = (\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)$$

Now, factor $$\sin^4 x - \cos^4 x$$ further as another difference of squares:

$$\sin^4 x - \cos^4 x = (\sin^2 x)^2 - (\cos^2 x)^2 = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)$$

Since $$\sin^2 x + \cos^2 x = 1$$, this simplifies to:

$$\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x) \cdot 1 = \sin^2 x - \cos^2 x$$

So the numerator becomes:

$$\sin^8 x - \cos^8 x = (\sin^2 x - \cos^2 x)(\sin^4 x + \cos^4 x)$$

Next, simplify $$\sin^4 x + \cos^4 x$$. We know:

$$\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$$

Substituting this back, the numerator is:

$$\sin^8 x - \cos^8 x = (\sin^2 x - \cos^2 x)(1 - 2\sin^2 x \cos^2 x)$$

Now, the original integral is:

$$\int \frac{(\sin^2 x - \cos^2 x)(1 - 2\sin^2 x \cos^2 x)}{1 - 2\sin^2 x \cos^2 x} dx$$

Notice that the denominator $$1 - 2\sin^2 x \cos^2 x$$ is the same as the second factor in the numerator. So, we can cancel this common factor (assuming it is not zero):

$$\int (\sin^2 x - \cos^2 x) dx$$

Recall the double-angle identity: $$\sin^2 x - \cos^2 x = -\cos 2x$$, because $$\cos 2x = \cos^2 x - \sin^2 x = -(\sin^2 x - \cos^2 x)$$. Therefore:

$$\int (\sin^2 x - \cos^2 x) dx = \int -\cos 2x dx$$

Now, integrate $$-\cos 2x$$:

$$-\int \cos 2x dx = -\frac{1}{2} \sin 2x + c$$

Because the integral of $$\cos kx$$ is $$\frac{\sin kx}{k}$$.

Thus, the integral simplifies to $$-\frac{1}{2} \sin 2x + c$$.

We must ensure that the denominator $$1 - 2\sin^2 x \cos^2 x$$ is never zero. Set it equal to zero:

$$1 - 2\sin^2 x \cos^2 x = 0 \implies 2\sin^2 x \cos^2 x = 1 \implies \sin^2 x \cos^2 x = \frac{1}{2}$$

Using $$\sin x \cos x = \frac{\sin 2x}{2}$$, so:

$$\left(\frac{\sin 2x}{2}\right)^2 = \frac{1}{2} \implies \frac{\sin^2 2x}{4} = \frac{1}{2} \implies \sin^2 2x = 2$$

But $$\sin^2 2x \leq 1$$, so $$\sin^2 2x = 2$$ has no real solution. Thus, the denominator is never zero, and the simplification is valid for all real $$x$$.

Comparing with the options:

A. $$-\frac{1}{2}\sin 2x + c$$

B. $$-\sin^2 x + c$$

C. $$-\frac{1}{2}\sin x + c$$

D. $$\frac{1}{2}\sin 2x + c$$

Option A matches our result.

Hence, the correct answer is Option A.