1. **Problem statement:** Calculate the definite integral $$\int_0^{20 \pi} |\sin x| \, dx$$. 2. **Formula and important rules:** The integral of the absolute value of sine over one period is the area under one full wave of $|\sin x|$. 3. The sine function has period $2\pi$, but $|\sin x|$ has period $\pi$ because $|\sin x| = |\sin(x+\pi)|$. 4. Calculate the integral over one period of $|\sin x|$ from $0$ to $\pi$: $$\int_0^{\pi} |\sin x| \, dx = \int_0^{\pi} \sin x \, dx = [-\cos x]_0^{\pi} = (-\cos \pi) - (-\cos 0) = (1) - (-1) = 2.$$ 5. Since the period of $|\sin x|$ is $\pi$, the integral from $0$ to $20\pi$ covers $20$ full periods. 6. Multiply the integral over one period by the number of periods: $$20 \times 2 = 40.$$ 7. **Final answer:** $$\int_0^{20 \pi} |\sin x| \, dx = 40.$$