The value of $\int_0^{20\pi} (\sin^4 x + \cos^4 x) \, dx$ is equal to:

Let $I = \int_0^{20\pi} (\sin^4 x + \cos^4 x) \, dx$

Using the algebraic identity $a^2 + b^2 = (a+b)^2 - 2ab$, the integrand can be simplified:

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

Since $\sin^2 x + \cos^2 x = 1$ and $2\sin x \cos x = \sin 2x$:

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

Using the half-angle formula $\sin^2 \theta = \dfrac{1 - \cos 2\theta}{2}$:

$\sin^4 x + \cos^4 x = 1 - \dfrac{1}{2} \left( \dfrac{1 - \cos 4x}{2} \right) = 1 - \dfrac{1}{4} + \dfrac{1}{4} \cos 4x = \dfrac{3}{4} + \dfrac{1}{4} \cos 4x$

Substituting this back into the integral:

$I = \int_0^{20\pi} \left( \dfrac{3}{4} + \dfrac{1}{4} \cos 4x \right) \, dx$

$I = \left[ \dfrac{3}{4} x + \dfrac{1}{16} \sin 4x \right]_0^{20\pi}$

Evaluating the limits:

$I = \left( \dfrac{3}{4} (20\pi) + \dfrac{1}{16} \sin(80\pi) \right) - (0 + 0)$

Since $\sin(80\pi) = 0$:

$I = 15\pi$

Answer: $15\pi$