Question

The value of the integral \[ \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin^4(4x)}{1 + e^{4x}} \, dx \] is equal to:

• A. \( \frac{3\pi}{128} \)
• B. \( \frac{3\pi}{256} \)
• C. \( \frac{3\pi}{64} \)
• D. \( \frac{3\pi}{32} \)

Hint:

The shortcut for \( \int_{-a}^{a} \frac{\text{Even}}{1 + \text{Exp}} \) is always the integral of the Even function from \( 0 \) to \( a \). This "kills" the messy denominator instantly.

Answer 1

The Correct Option is C

Concept: This problem is efficiently solved using the "King's Property" of definite integrals combined with Wallis' Formula for reduction:
• King's Property: \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \).
• Symmetry: For an integral \( \int_{-a}^{a} \frac{f(x)}{1 + e^{kx}} \, dx \), if \( f(x) \) is an even function, the result simplifies to \( \int_{0}^{a} f(x) \, dx \).
• Wallis' Formula: A shortcut to evaluate \( \int_{0}^{\pi/2} \sin^n \theta \, d\theta \).

Step 1: Simplifying the integral using symmetry properties.
Let \( I = \int_{-\pi/8}^{\pi/8} \frac{\sin^4(4x)}{1 + e^{4x}} \, dx \). Since \( \sin^4(4x) \) is an even function, applying the property \( x \to -x \) and adding the integrals cancels the exponential term: \[ 2I = \int_{-\pi/8}^{\pi/8} \sin^4(4x) \, dx \quad \Rightarrow \quad I = \int_{0}^{\pi/8} \sin^4(4x) \, dx \]

Step 2: Changing the variable for Wallis' Formula.
To use Wallis' Formula, we need the upper limit to be \( \pi/2 \). Let \( 4x = \theta \), then \( 4dx = d\theta \Rightarrow dx = \frac{d\theta}{4} \).
• When \( x = 0, \theta = 0 \).
• When \( x = \pi/8, \theta = \pi/2 \). Substituting into the integral: \[ I = \int_{0}^{\pi/2} \sin^4 \theta \cdot \frac{d\theta}{4} = \frac{1}{4} \int_{0}^{\pi/2} \sin^4 \theta \, d\theta \]

Step 3: Applying Wallis' Formula.
For \( n = 4 \) (even): \[ \int_{0}^{\pi/2} \sin^4 \theta \, d\theta = \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{3\pi}{16} \] Now, calculate the final value of \( I \): \[ I = \frac{1}{4} \times \frac{3\pi}{16} = \frac{3\pi}{64} \]