int_{0}^{pi} sqrt{frac{1 + cos 2x}{2}} , dx i | vedclass

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(B) We know that $1 + \cos 2x = 2 \cos^2 x$. Substituting this into the integral,we get: $I = \int_{0}^{\pi} \sqrt{\frac{2 \cos^2 x}{2}} \, dx = \int_

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$2$

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(B) We know that $1 + \cos 2x = 2 \cos^2 x$. Substituting this into the integral,we get: $I = \int_{0}^{\pi} \sqrt{\frac{2 \cos^2 x}{2}} \, dx = \int_{0}^{\pi} |\cos x| \, dx$. Since $\cos x \ge 0$ for $x \in [0, \pi/2]$ and $\cos x \le 0$ for $x \in [\pi/2, \pi]$,we split the integral: $I = \int_{0}^{\pi/2} \cos x \, dx - \int_{\pi/2}^{\pi} \cos x \, dx$. Evaluating the integrals: $I = [\sin x]_{0}^{\pi/2} - [\sin x]_{\pi/2}^{\pi}$. $I = (\sin(\pi/2) - \sin(0)) - (\sin(\pi) - \sin(\pi/2))$. $I = (1 - 0) - (0 - 1) = 1 + 1 = 2$.

$\int_{0}^{\pi} \sqrt{\frac{1 + \cos 2x}{2}} \, dx$ is equal to

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