Question

Evaluate the integral: \[ I = \int_{0}^{32\pi} \sqrt{1 - \cos 4x} \,dx \]

• A. \( 16\sqrt{2} \)
• B. \( 32\sqrt{2} \)
• C. \( 128\sqrt{2} \)
• D. \( 64\sqrt{2} \)

Hint:

When integrating functions involving absolute values of sine or cosine, break the integral into periodic segments where the function maintains a consistent sign.

Answer 1

The Correct Option is D

Step 1: Using the standard trigonometric identity. We use the identity: \[ 1 - \cos 4x = 2\sin^2 2x \] Thus, the given integral transforms as: \[ I = \int_{0}^{32\pi} \sqrt{2\sin^2 2x} \,dx \] \[ I = \sqrt{2} \int_{0}^{32\pi} |\sin 2x| \,dx \] Step 2: Evaluating the absolute value integral. Since \(\sin 2x\) is a periodic function with period \(\pi\), we break it into intervals where \(\sin 2x\) is positive and negative. The function \(\sin 2x\) completes one full cycle in the interval \( [0, \pi] \) and repeats. Since the total integration range is \( 32\pi \), we have: \[ \frac{32\pi}{\pi} = 32 \text{ complete cycles.} \] Each complete cycle contributes: \[ \int_{0}^{\pi} |\sin 2x| \,dx = 2 \] Thus, the total integral is: \[ I = \sqrt{2} \times (32 \times 2) = 64\sqrt{2} \] Final Answer: \[ \boxed{64\sqrt{2}} \]