If $$S = \left\{x \in \mathbb{R} : \sin^{-1}\left(\frac{x+1}{\sqrt{x^2+2x+2}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}\right\}$$ then $$\sum_{x \in S}\left(\sin\left((x^2+x+5)\frac{\pi}{2}\right) - \cos\left((x^2+x+5)\pi\right)\right)$$ is equal to _____.

Given: $$\sin^{-1}\left(\frac{x+1}{\sqrt{x^2+2x+2}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}$$

Note that $$x^2 + 2x + 2 = (x+1)^2 + 1$$.

Substitution: Let $$\tan\theta_1 = x + 1$$ and $$\tan\theta_2 = x$$.

Then $$\sin\theta_1 = \frac{x+1}{\sqrt{(x+1)^2+1}}$$ and $$\sin\theta_2 = \frac{x}{\sqrt{x^2+1}}$$.

The equation becomes $$\theta_1 - \theta_2 = \frac{\pi}{4}$$, so:

$$\tan(\theta_1 - \theta_2) = \frac{\tan\theta_1 - \tan\theta_2}{1 + \tan\theta_1\tan\theta_2} = \frac{(x+1) - x}{1 + x(x+1)} = \frac{1}{1 + x^2 + x}$$

Setting this equal to $$\tan\frac{\pi}{4} = 1$$:

$$\frac{1}{1 + x + x^2} = 1 \implies x^2 + x = 0 \implies x(x+1) = 0$$

So $$x = 0$$ or $$x = -1$$.

Verification:

At $$x = 0$$: $$\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) - \sin^{-1}(0) = \frac{\pi}{4}$$ ✓

At $$x = -1$$: $$\sin^{-1}(0) - \sin^{-1}\left(\frac{-1}{\sqrt{2}}\right) = 0 + \frac{\pi}{4} = \frac{\pi}{4}$$ ✓

$$S = \{0, -1\}$$

Computing the sum:

For both $$x = 0$$ and $$x = -1$$: $$x^2 + x + 5 = 5$$.

$$\sin\left(\frac{5\pi}{2}\right) - \cos(5\pi) = \sin\left(\frac{\pi}{2}\right) - \cos(\pi) = 1 - (-1) = 2$$

Total sum = $$2 + 2 = 4$$.

The answer is $$4$$.