int_{0}^{pi/4} sqrt{1+sin 2x} dx = rule{1cm}{ | vedclass

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(B) We know that $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$.Therefore,$\sqrt{1+\sin 2x} = \sqrt{\sin^2 x + \cos^2 x + 2 \sin x \cos x}

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$1$

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(B) We know that $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$.Therefore,$\sqrt{1+\sin 2x} = \sqrt{\sin^2 x + \cos^2 x + 2 \sin x \cos x} = \sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x|$.Since $x \in [0, \pi/4]$,both $\sin x$ and $\cos x$ are non-negative,so $|\sin x + \cos x| = \sin x + \cos x$.Now,the integral becomes $\int_{0}^{\pi/4} (\sin x + \cos x) dx$.Integrating term by term,we get $[-\cos x + \sin x]_0^{\pi/4}$.Evaluating at the limits: $(-\cos(\pi/4) + \sin(\pi/4)) - (-\cos(0) + \sin(0))$.$= (-1/\sqrt{2} + 1/\sqrt{2}) - (-1 + 0) = 0 - (-1) = 1$.

$\int_{0}^{\pi/4} \sqrt{1+\sin 2x} dx = \rule{1cm}{0.15mm}$

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