Let I = int_{0}^{100 pi} sqrt{1 - cos 2x} , d | vedclass

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(B) Given $I = \int_{0}^{100 \pi} \sqrt{1 - \cos 2x} \, dx$.Using the identity $1 - \cos 2x = 2 \sin^2 x$, we have $I = \int_{0}^{100 \pi} \sqrt{2 \si

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$I = 200 \sqrt{2}$

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(B) Given $I = \int_{0}^{100 \pi} \sqrt{1 - \cos 2x} \, dx$.Using the identity $1 - \cos 2x = 2 \sin^2 x$, we have $I = \int_{0}^{100 \pi} \sqrt{2 \sin^2 x} \, dx$.$I = \sqrt{2} \int_{0}^{100 \pi} |\sin x| \, dx$.Since $|\sin x|$ is a periodic function with period $\pi$, we can write $I = \sqrt{2} 	imes 100 \int_{0}^{\pi} |\sin x| \, dx$.In the interval $[0, \pi]$, $\sin x \geq 0$, so $|\sin x| = \sin x$.$I = 100 \sqrt{2} \int_{0}^{\pi} \sin x \, dx$.$I = 100 \sqrt{2} [-\cos x]_{0}^{\pi}$.$I = 100 \sqrt{2} [-\cos \pi - (-\cos 0)]$.$I = 100 \sqrt{2} [-(-1) - (-1)]$.$I = 100 \sqrt{2} [1 + 1] = 100 \sqrt{2} 	imes 2 = 200 \sqrt{2}$.

Let $I = \int_{0}^{100 \pi} \sqrt{1 - \cos 2x} \, dx$, then

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