Question

Evaluate the integral: \[ I = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin^4(4x)}{1 + e^{4x}} \, dx \]

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

Hint:

For integrals involving powers of trigonometric functions, always start by simplifying the power using identities like \( \sin^2 x = \frac{1 - \cos(2x)}{2} \). Break the integral into simpler terms and use standard integration formulas for trigonometric and exponential functions.

Answer 1

The Correct Option is C

We need to evaluate the integral: \[ I = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin^4(4x)}{1 + e^{4x}} \, dx. \] --- Step 1: Apply Symmetry Property of Definite Integrals We use the property: \[ \int_{-a}^{a} f(x) \, dx = \int_{-a}^{a} f(-x) \, dx. \] Substituting \( x \to -x \) in the given integral: \[ I = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin^4(4(-x))}{1 + e^{4(-x)}} \, dx. \] Since \( \sin(-\theta) = -\sin \theta \), we get: \[ \sin^4(4(-x)) = \sin^4(4x). \] Also, since \( e^{-4x} = \frac{1}{e^{4x}} \), we rewrite the denominator: \[ 1 + e^{-4x} = \frac{e^{4x} + 1}{e^{4x}}. \] Thus, the transformed integral is: \[ I = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin^4(4x)}{1 + e^{-4x}} \, dx = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin^4(4x) e^{4x}}{e^{4x} + 1} \, dx. \] Adding the original integral and the transformed integral: \[ 2I = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \sin^4(4x) \, dx. \] Thus, we get: \[ I = \frac{1}{2} \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \sin^4(4x) \, dx. \] --- Step 2: Solve the Integral \( \int \sin^4(4x) \, dx \) Using the identity: \[ \sin^4 A = \frac{3}{8} - \frac{1}{2} \cos 8A + \frac{1}{8} \cos 16A. \] \[ \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \sin^4(4x) \, dx = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \left( \frac{3}{8} - \frac{1}{2} \cos 32x + \frac{1}{8} \cos 64x \right) dx. \] Since \( \int_{-a}^{a} \cos kx \, dx = 0 \), the integral simplifies to: \[ \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{3}{8} dx. \] \[ = \frac{3}{8} \times \frac{\pi}{4} = \frac{3\pi}{32}. \] Thus, \[ I = \frac{1}{2} \times \frac{3\pi}{32} = \frac{3\pi}{64}. \] --- Final Answer: \[ \boxed{\frac{3\pi}{64}} \] which matches option (3). \bigskip