$$\displaystyle\int_0^{20\pi} (|\sin x| + |\cos x|)^2 dx$$ is equal to:

We need to evaluate $$\displaystyle\int_0^{20\pi} (|\sin x| + |\cos x|)^2\, dx$$.

Expanding: $$(|\sin x| + |\cos x|)^2 = \sin^2 x + 2|\sin x||\cos x| + \cos^2 x = 1 + 2|\sin x \cos x|$$.

Using the identity $$2\sin x \cos x = \sin 2x$$, we get $$2|\sin x \cos x| = |\sin 2x|$$.

So our integral becomes $$\displaystyle\int_0^{20\pi} (1 + |\sin 2x|)\, dx = \int_0^{20\pi} 1\, dx + \int_0^{20\pi} |\sin 2x|\, dx$$.

The first integral is simply $$20\pi$$.

For the second integral, $$|\sin 2x|$$ has period $$\dfrac{\pi}{2}$$. The number of complete periods in $$[0, 20\pi]$$ is $$\dfrac{20\pi}{\pi/2} = 40$$.

Over one period $$\left[0, \dfrac{\pi}{2}\right]$$, $$\sin 2x \geq 0$$, so $$|\sin 2x| = \sin 2x$$.

$$\int_0^{\pi/2} \sin 2x\, dx = \left[-\dfrac{\cos 2x}{2}\right]_0^{\pi/2} = -\dfrac{\cos \pi}{2} + \dfrac{\cos 0}{2} = \dfrac{1}{2} + \dfrac{1}{2} = 1$$

Therefore, $$\displaystyle\int_0^{20\pi} |\sin 2x|\, dx = 40 \times 1 = 40$$.

Total = $$20\pi + 40 = 20(\pi + 2)$$.

The answer is Option D: $$20(\pi + 2)$$.