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(D) Given the integral: $I = \int \sqrt{1 - \sin 2x} \, dx$Using the trigonometric identity $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$,

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$\sin x + \cos x + c$

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(D) Given the integral: $I = \int \sqrt{1 - \sin 2x} \, dx$Using the trigonometric identity $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$,we have:$1 - \sin 2x = \sin^2 x + \cos^2 x - 2 \sin x \cos x = (\cos x - \sin x)^2$Thus,$\sqrt{1 - \sin 2x} = \sqrt{(\cos x - \sin x)^2} = |\cos x - \sin x|$Since $x \in (0, \pi/4)$,we know that $\cos x > \sin x$,so $|\cos x - \sin x| = \cos x - \sin x$Now,integrating the expression:$I = \int (\cos x - \sin x) \, dx$$I = \sin x - (-\cos x) + c$$I = \sin x + \cos x + c$

$\int \sqrt{1 - \sin 2x} \, dx = \dots \dots, \text{ where } x \in (0, \pi/4)$

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