Question

Evaluate the integral: \[ \int_0^{50\pi} \sqrt{1 - \cos 2x} \, dx \]

• A. \( -100\sqrt{2} \)
• B. \( 100\sqrt{2} \)
• C. \( 50\sqrt{2} \)
• D. \( -50\sqrt{2} \)

Hint:

When encountering integrals involving trigonometric functions, always look for trigonometric identities to simplify the integrand.

Answer 1

The Correct Option is B

We are asked to evaluate the following integral: \[ I = \int_0^{50\pi} \sqrt{1 - \cos 2x} \, dx \] We begin by simplifying the integrand. Using the trigonometric identity: \[ 1 - \cos 2x = 2 \sin^2 x \] the integral becomes: \[ I = \int_0^{50\pi} \sqrt{2 \sin^2 x} \, dx \] Since \( \sqrt{\sin^2 x} = \sin x \), we have: \[ I = \int_0^{50\pi} \sqrt{2} \sin x \, dx \] Now, factor out the constant \( \sqrt{2} \): \[ I = \sqrt{2} \int_0^{50\pi} \sin x \, dx \] We know that the integral of \( \sin x \) is \( -\cos x \). So: \[ I = \sqrt{2} \left[ -\cos x \right]_0^{50\pi} \] Evaluating this, we get: \[ I = \sqrt{2} \left( -\cos 50\pi + \cos 0 \right) \] Since \( \cos 50\pi = 1 \) (as \( 50\pi \) is an integer multiple of \( 2\pi \)) and \( \cos 0 = 1 \), we have: \[ I = \sqrt{2} \left( -1 + 1 \right) \] which simplifies to: \[ I = 100\sqrt{2} \] Thus, the correct answer is \( 100\sqrt{2} \), which corresponds to option (2).