int_0^{pi / 4} sqrt{1-sin 2 x} ,d x = | vedclass

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(C) We know that $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$.So,$1 - \sin 2x = \sin^2 x + \cos^2 x - 2 \sin x \cos x = (\cos x - \sin x)

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$\sqrt{2}-1$

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(C) We know that $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$.So,$1 - \sin 2x = \sin^2 x + \cos^2 x - 2 \sin x \cos x = (\cos x - \sin x)^2$.The integral becomes $\int_0^{\pi / 4} \sqrt{(\cos x - \sin x)^2} \,d x = \int_0^{\pi / 4} |\cos x - \sin x| \,d x$.In the interval $[0, \pi / 4]$,$\cos x \geq \sin x$,so $|\cos x - \sin x| = \cos x - \sin x$.Thus,the integral is $\int_0^{\pi / 4} (\cos x - \sin x) \,d x = [\sin x + \cos x]_0^{\pi / 4}$.Evaluating the limits: $(\sin(\pi / 4) + \cos(\pi / 4)) - (\sin 0 + \cos 0) = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1$.

$\int_0^{\pi / 4} \sqrt{1-\sin 2 x} \,d x =$

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