Question

\( \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 \theta}{1 + e^{\sin \theta}} \, d\theta \) is equal to:

• A. \( 3\pi + 8 \)
• B. \( 3\pi + 4 \)
• C. \( 4\pi + 8 \)
• D. \( 4\pi + 4 \)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a definite integral with symmetric limits \([-a, a]\). We use King's Property: \(\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx\). For symmetric limits, this property helps eliminate the exponential term in the denominator.
Step 2: Key Formula or Approach:
1. Let \( I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 \theta}{1 + e^{\sin \theta}} \, d\theta \).
2. Apply King's property: Replace \( \theta \) with \( (-\pi/4 + \pi/4 - \theta) = -\theta \).
3. Add the two expressions of \( I \).
Step 3: Detailed Explanation:
1. \( I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 \theta}{1 + e^{\sin \theta}} \, d\theta \dots \text{(Eq. 1)} \)
2. Replacing \( \theta \) with \( -\theta \) (since \( \cos(-\theta) = \cos \theta \) and \( \sin(-\theta) = -\sin \theta \)): \[ I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 \theta}{1 + e^{-\sin \theta}} \, d\theta = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 \theta \cdot e^{\sin \theta}}{e^{\sin \theta} + 1} \, d\theta \dots \text{(Eq. 2)} \] 3. Adding Eq. 1 and Eq. 2: \[ 2I = \int_{-\pi/4}^{\pi/4} 32 \cos^4 \theta \left( \frac{1 + e^{\sin \theta}}{1 + e^{\sin \theta}} \right) \, d\theta = 32 \int_{-\pi/4}^{\pi/4} \cos^4 \theta \, d\theta \] 4. Since \( \cos^4 \theta \) is an even function: \[ 2I = 64 \int_{0}^{\pi/4} \cos^4 \theta \, d\theta \implies I = 32 \int_{0}^{\pi/4} \left( \frac{1 + \cos 2\theta}{2} \right)^2 \, d\theta \] \[ I = 8 \int_{0}^{\pi/4} (1 + 2\cos 2\theta + \cos^2 2\theta) \, d\theta = 8 \int_{0}^{\pi/4} \left( 1 + 2\cos 2\theta + \frac{1 + \cos 4\theta}{2} \right) \, d\theta \] \[ I = 4 \int_{0}^{\pi/4} (3 + 4\cos 2\theta + \cos 4\theta) \, d\theta \] 5. Evaluating the integral: \[ I = 4 \left[ 3\theta + 2\sin 2\theta + \frac{\sin 4\theta}{4} \right]_0^{\pi/4} = 4 \left[ \left( \frac{3\pi}{4} + 2\sin \frac{\pi}{2} + \frac{\sin \pi}{4} \right) - 0 \right] \] \[ I = 4 \left( \frac{3\pi}{4} + 2 \right) = 3\pi + 8 \]
Step 4: Final Answer:
The value of the integral is \( 3\pi + 8 \).