Evaluate the integral: int_0^{50 pi} sqrt{1-c | vedclass

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(B) We know that $1 - \cos 2x = 2 \sin^2 x$. Substituting this into the integral,we get: $I = \int_0^{50 \pi} \sqrt{2 \sin^2 x} \, dx = \sqrt{2} \int_

Vedclass · Accepted Answer

$100$

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(B) We know that $1 - \cos 2x = 2 \sin^2 x$. Substituting this into the integral,we get: $I = \int_0^{50 \pi} \sqrt{2 \sin^2 x} \, dx = \sqrt{2} \int_0^{50 \pi} |\sin x| \, dx$. Since $|\sin x|$ is a periodic function with period $\pi$,we can write: $I = \sqrt{2} 	imes 50 \int_0^{\pi} |\sin x| \, dx$. In the interval $[0, \pi]$,$\sin x \ge 0$,so $|\sin x| = \sin x$. $I = 50 \sqrt{2} \int_0^{\pi} \sin x \, dx$. Evaluating the integral: $I = 50 \sqrt{2} [-\cos x]_0^{\pi} = 50 \sqrt{2} (-(\cos \pi - \cos 0)) = 50 \sqrt{2} (-(-1 - 1)) = 50 \sqrt{2} (2) = 100 \sqrt{2}$.

Evaluate the integral: $\int_0^{50 \pi} \sqrt{1-\cos 2x} \, dx$ (in $\sqrt{2}$)

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