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Let $I=\int_{0}^{{100 \pi }} \sqrt{(1-\cos 2 x)} d x$, then
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Verified Answer
The correct answer is:
$I=200 \sqrt{2}$
$I=\int_{0}^{1004} \sqrt{1-\cos 2 x} d x$
$=\int_{0}^{100 \pi} \sqrt{2 \sin ^{2} x} d x$
$=\sqrt{2} \int_{0}^{100 \pi} \sin x \mid d x$
$=\sqrt{2} \times 100 \int_{0}^{n} \sin x|d x|$
$[\sin x \mid$ has period of $\pi]$
$=100 \sqrt{2} \int_{0}^{\pi} sin x d x$
$=100 \sqrt{2}[-\cos x]_{0}^{\pi}$
$=100 \sqrt{2}\{(-\cos \pi)-(-\cos 0)]$
$=100 \sqrt{2}[-(-1)-(-1)]$
$=100 \sqrt{2} \times 2$
$=200 \sqrt{2}$
$=\int_{0}^{100 \pi} \sqrt{2 \sin ^{2} x} d x$
$=\sqrt{2} \int_{0}^{100 \pi} \sin x \mid d x$
$=\sqrt{2} \times 100 \int_{0}^{n} \sin x|d x|$
$[\sin x \mid$ has period of $\pi]$
$=100 \sqrt{2} \int_{0}^{\pi} sin x d x$
$=100 \sqrt{2}[-\cos x]_{0}^{\pi}$
$=100 \sqrt{2}\{(-\cos \pi)-(-\cos 0)]$
$=100 \sqrt{2}[-(-1)-(-1)]$
$=100 \sqrt{2} \times 2$
$=200 \sqrt{2}$
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