int_0^{400 pi} sqrt{1-cos 2 x} , dx = (in sqr | vedclass

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(D) We know that $1 - \cos 2x = 2 \sin^2 x$. Therefore,$\sqrt{1 - \cos 2x} = \sqrt{2 \sin^2 x} = \sqrt{2} |\sin x|$. The integral becomes $I = \int_0^

Vedclass · Accepted Answer

$800$

Vedclass · Answer

(D) We know that $1 - \cos 2x = 2 \sin^2 x$. Therefore,$\sqrt{1 - \cos 2x} = \sqrt{2 \sin^2 x} = \sqrt{2} |\sin x|$. The integral becomes $I = \int_0^{400 \pi} \sqrt{2} |\sin x| \, dx$. Since $|\sin x|$ is periodic with period $\pi$,we can write $I = \sqrt{2} 	imes 400 \int_0^{\pi} |\sin x| \, dx$. In the interval $[0, \pi]$,$\sin x \ge 0$,so $|\sin x| = \sin x$. Thus,$I = 400 \sqrt{2} \int_0^{\pi} \sin x \, dx$. Evaluating the integral: $\int_0^{\pi} \sin x \, dx = [-\cos x]_0^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$. Finally,$I = 400 \sqrt{2} 	imes 2 = 800 \sqrt{2}$.

$\int_0^{400 \pi} \sqrt{1-\cos 2 x} \, dx =$ (in $\sqrt{2}$)

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