The value of the integral $$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{32\cos^4 x}{1 + e^{\sin x}}\right)dx$$ is:

$$I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 x}{1 + e^{\sin x}} \, dx \quad \text{--- (Equation 1)}$$

$$I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4(-x)}{1 + e^{\sin(-x)}} \, dx$$

$$I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 x \cdot e^{\sin x}}{e^{\sin x} + 1} \, dx \quad \text{--- (Equation 2)}$$

$$2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 x \left(1 + e^{\sin x}\right)}{1 + e^{\sin x}} \, dx$$

$$2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 32 \cos^4 x \, dx$$

$$I = 16 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^4 x \, dx$$

$$I = 32 \int_{0}^{\frac{\pi}{4}} \cos^4 x \, dx$$

$$\cos^4 x = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1 + 2\cos 2x + \cos^2 2x}{4}$$

$$\cos^4 x = \frac{1}{4} \left( 1 + 2\cos 2x + \frac{1 + \cos 4x}{2} \right) = \frac{1}{8} (3 + 4\cos 2x + \cos 4x)$$

$$I = 32 \int_{0}^{\frac{\pi}{4}} \frac{1}{8} (3 + 4\cos 2x + \cos 4x) \, dx$$

$$I = 4 \left[ 3x + 2\sin 2x + \frac{\sin 4x}{4} \right]_{0}^{\frac{\pi}{4}}$$

$$I = 4 \left( \frac{3\pi}{4} + 2(1) + 0 \right)$$

$$I = 3\pi + 8$$