Step 1: Understanding the Concept:
This problem highlights the utility of the "King's Property" of definite integrals, which states that \(\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx\).
In this specific case, the interval is symmetric \([- \pi/8, \pi/8]\), so \(a+b = 0\).
Replacing \(x\) with \(-x\) is a powerful technique for integrals involving exponential denominators like \(1+e^x\).
Because \(\sin^4(4x)\) is an even function, substituting \(-x\) doesn't change the numerator but flips the sign in the exponent of the denominator.
Adding the original integral and the transformed integral often eliminates the messy exponential term entirely.
This is a standard "trick" in advanced integration problems where symmetry is present but obscured by a non-symmetric denominator.
Step 2: Key Formula or Approach:
1. King's Property: \(I = \int_{a}^{b} f(a+b-x) dx\).
2. Property for even \(f(x)\): \(\int_{-a}^{a} \frac{f(x)}{1+e^{kx}} dx = \int_{0}^{a} f(x) dx\).
3. Wallis' formula for \(\sin^n \theta\).
Step 3: Detailed Explanation:
Let \(I = \int_{-\pi/8}^{\pi/8} \frac{\sin^4(4x)}{1+e^{4x}} dx\).
Apply \(x \to -x\):
\[ I = \int_{-\pi/8}^{\pi/8} \frac{\sin^4(4(-x))}{1+e^{4(-x)}} dx = \int_{-\pi/8}^{\pi/8} \frac{\sin^4(4x)}{1+e^{-4x}} dx = \int_{-\pi/8}^{\pi/8} \frac{e^{4x} \sin^4(4x)}{e^{4x}+1} dx \]
Adding the two expressions for \(I\):
\[ 2I = \int_{-\pi/8}^{\pi/8} \left( \frac{\sin^4(4x)}{1+e^{4x}} + \frac{e^{4x} \sin^4(4x)}{1+e^{4x}} \right) dx = \int_{-\pi/8}^{\pi/8} \frac{\sin^4(4x)(1+e^{4x})}{1+e^{4x}} dx \]
The exponential term cancels out perfectly:
\[ 2I = \int_{-\pi/8}^{\pi/8} \sin^4(4x) dx \]
Since \(\sin^4(4x)\) is an even function: \(2I = 2 \int_{0}^{\pi/8} \sin^4(4x) dx \implies I = \int_{0}^{\pi/8} \sin^4(4x) dx\).
Let \(4x = \theta\), then \(4dx = d\theta\). Limits change from \([0, \pi/8]\) to \([0, \pi/2]\).
\[ I = \frac{1}{4} \int_{0}^{\pi/2} \sin^4 \theta d\theta \]
Applying Wallis' formula for \(n=4\):
\[ \text{Value} = \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} = \frac{3\pi}{16} \]
Calculating final result: \(I = \frac{1}{4} \cdot \frac{3\pi}{16} = \frac{3\pi}{64}\).
Step 4: Final Answer:
The definite integral evaluates to \(\frac{3\pi}{64}\).