Question

The value of the integral \( \int_{0}^{32\pi} \sqrt{1 - \cos 4x} \, dx \) is equal to:

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

Hint:

The integral of \( |\sin kx| \) or \( |\cos kx| \) over one half-period is always \( \frac{2}{k} \). For \( |\sin 2x| \), the integral over \( [0, \pi/2] \) is \( \frac{2}{2} = 1 \).

Answer 1

The Correct Option is C

Concept: The solution involves trigonometric simplification and the property of periodic functions in definite integrals:
• Half-angle formula: \( 1 - \cos \theta = 2\sin^2(\frac{\theta}{2}) \).
• Identity: \( \sqrt{x^2} = |x| \).
• Periodic Property: If \( f(x) \) is periodic with period \( T \), then \( \int_{0}^{nT} f(x) \, dx = n \int_{0}^{T} f(x) \, dx \).

Step 1: Simplifying the integrand using trigonometric identities.
We know that \( 1 - \cos 4x = 2\sin^2(2x) \). Substituting this into the integral: \[ I = \int_{0}^{32\pi} \sqrt{2\sin^2(2x)} \, dx = \sqrt{2} \int_{0}^{32\pi} |\sin 2x| \, dx \]

Step 2: Applying the periodicity property.
The function \( f(x) = |\sin 2x| \) is periodic. The period of \( \sin x \) is \( 2\pi \), so the period of \( \sin 2x \) is \( \frac{2\pi}{2} = \pi \).
However, the period of the absolute value function \( |\sin 2x| \) is \( \frac{\pi}{2} \).
We can express the upper limit \( 32\pi \) in terms of the period \( \frac{\pi}{2} \): \[ 32\pi = 64 \times \left( \frac{\pi}{2} \right) \] Using the property \( \int_{0}^{nT} f(x) \, dx = n \int_{0}^{T} f(x) \, dx \): \[ I = \sqrt{2} \times 64 \int_{0}^{\pi/2} |\sin 2x| \, dx \]

Step 3: Evaluating the definite integral.
In the interval \( [0, \pi/2] \), \( 2x \) ranges from \( 0 \) to \( \pi \). In this range, \( \sin 2x \) is non-negative, so we can remove the modulus: \[ I = 64\sqrt{2} \int_{0}^{\pi/2} \sin 2x \, dx \] \[ I = 64\sqrt{2} \left[ \frac{-\cos 2x}{2} \right]_{0}^{\pi/2} = \frac{64\sqrt{2}}{2} [ -\cos(\pi) - (-\cos 0) ] \] \[ I = 32\sqrt{2} [ -(-1) + 1 ] = 32\sqrt{2} [ 1 + 1 ] = 64\sqrt{2} \times 2 = 128\sqrt{2} \]