Question

Find \( I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 \theta}{1 + e^{\sin \theta}} \, d\theta \)

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

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We use the property of definite integrals: \( \int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx \). For a symmetric interval \([-a, a]\), this simplifies to \( \int_{-a}^a f(x) \, dx = \int_{-a}^a f(-x) \, dx \). Adding these two forms often helps eliminate exponential terms like \( e^{\sin \theta} \).

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. Using the property, \( I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4(-\theta)}{1 + e^{\sin(-\theta)}} \, d\theta = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 \theta}{1 + e^{-\sin \theta}} \, d\theta \).

Step 3: Detailed Explanation:
1. Add the two expressions for \( I \): \[ 2I = \int_{-\pi/4}^{\pi/4} 32 \cos^4 \theta \left( \frac{1}{1 + e^{\sin \theta}} + \frac{1}{1 + e^{-\sin \theta}} \right) d\theta \] 2. Note that \( \frac{1}{1+e^x} + \frac{1}{1+e^{-x}} = 1 \). Thus: \[ 2I = \int_{-\pi/4}^{\pi/4} 32 \cos^4 \theta \, d\theta \implies I = 16 \int_{-\pi/4}^{\pi/4} \cos^4 \theta \, d\theta \] 3. Since \( \cos^4 \theta \) is an even function: \[ I = 32 \int_{0}^{\pi/4} \cos^4 \theta \, d\theta = 32 \int_{0}^{\pi/4} \left( \frac{1 + \cos 2\theta}{2} \right)^2 d\theta \] \[ I = 8 \int_{0}^{\pi/4} (1 + 2\cos 2\theta + \cos^2 2\theta) \, d\theta = 8 \int_{0}^{\pi/4} \left( 1 + 2\cos 2\theta + \frac{1 + \cos 4\theta}{2} \right) d\theta \] \[ I = 4 \int_{0}^{\pi/4} (3 + 4\cos 2\theta + \cos 4\theta) \, d\theta = 4 \left[ 3\theta + 2\sin 2\theta + \frac{\sin 4\theta}{4} \right]_0^{\pi/4} \] \[ I = 4 \left( \frac{3\pi}{4} + 2(1) + 0 \right) = 3\pi + 8 \]

Step 4: Final Answer:
The value of the integral is \( 3\pi + 8 \).