If $$\int_0^{\pi} \frac{5^{\cos x}(1+\cos x \cos 3x+\cos^2 x+\cos^3 x \cos 3x)dx}{1+5^{\cos x}} = \frac{k\pi}{16}$$, then $$k$$ is equal to _____.

$$I = \int_{0}^{\pi} \frac{5^{\cos x} \left( 1 + \cos x \cos 3x + \cos^2 x + \cos^3 x \cos 3x \right)}{1 + 5^{\cos x}} \, dx$$

Applying King's Property ($$x \to \pi - x$$):

$$I = \int_{0}^{\pi} \frac{5^{-\cos x} \left( 1 + \cos x \cos 3x + \cos^2 x + \cos^3 x \cos 3x \right)}{1 + 5^{-\cos x}} \, dx$$

$$2I = \int_{0}^{\pi} \left( 1 + \cos x \cos 3x + \cos^2 x + \cos^3 x \cos 3x \right) dx$$

$$2I = 2 \int_{0}^{\pi/2} \left( 1 + \cos x \cos 3x + \cos^2 x + \cos^3 x \cos 3x \right) dx$$

$$I = \int_{0}^{\pi/2} \left( 1 + \sin x(-\sin 3x) + \sin^2 x - \sin^3 x \sin 3x \right) dx$$

$$2I = \int_{0}^{\pi/2} \left( 3 + \cos 4x + \cos^3 x \cos 3x - \sin^3 x \sin 3x \right) dx$$

$$2I = \int_{0}^{\pi/2} \left( 3 + \cos 4x + \left[\frac{\cos 3x + 3\cos x}{4}\right]\cos 3x - \sin 3x\left[\frac{3\sin x - \sin 3x}{4}\right] \right) dx$$

$$2I = \int_{0}^{\pi/2} \left( 3 + \cos 4x + \frac{1}{4} + \frac{3}{4}\cos 4x \right) dx$$

$$2I = \left[ \frac{13}{4}x + \frac{7}{16}\sin 4x \right]_{0}^{\pi/2} \implies 2I = \frac{13}{4} \cdot \frac{\pi}{2} \implies I = \frac{13\pi}{16}$$