Question

Evaluate: \( \int_0^{400\pi} \sqrt{1 - \cos 2x} \, dx \)

• A. \( 100\sqrt{2} \)
• B. \( 200\sqrt{2} \)
• C. \( 400\sqrt{2} \)
• D. \( 800\sqrt{2} \)

Hint:

When you see \( \sqrt{1 - \cos 2x} \), always try using the identity \( 1 - \cos 2x = 2 \sin^2 x \), and simplify using absolute value if needed.

Answer 1

The Correct Option is D

Step 1: Simplify the integrand using identity We know the identity: \[ 1 - \cos 2x = 2 \sin^2 x \] So, \[ \sqrt{1 - \cos 2x} = \sqrt{2 \sin^2 x} = \sqrt{2} |\sin x| \] Step 2: Rewrite the integral
\[ \int_0^{400\pi} \sqrt{1 - \cos 2x} \, dx = \int_0^{400\pi} \sqrt{2} |\sin x| \, dx = \sqrt{2} \int_0^{400\pi} |\sin x| \, dx \] Step 3: Use periodicity of \( |\sin x| \)
The function \( |\sin x| \) is periodic with period \( \pi \). Over each interval of length \( \pi \), we have: \[ \int_0^{\pi} |\sin x| \, dx = \int_0^{\pi} \sin x \, dx = 2 \] Step 4: Calculate total contribution over 400 periods \[ \int_0^{400\pi} |\sin x| \, dx = 400 \cdot 2 = 800 \] Step 5: Multiply by \( \sqrt{2} \) \[ \int_0^{400\pi} \sqrt{1 - \cos 2x} \, dx = \sqrt{2} \cdot 800 = 800\sqrt{2} \]