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$\int_0^{\pi / 4} \sqrt{1-\sin 2 x} d x=$
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$\sqrt{2}-1$
$\begin{aligned} & \int_0^{\pi / 4} \sqrt{1-\sin 2 x} \mathrm{~d} x=\int_0^{\pi / 4} \sqrt{\cos ^2 x+\sin ^2 x-2 \sin x \cdot \cos x} \mathrm{~d} x \\ & =\int_0^{\pi / 4} \sqrt{(\cos x-\sin x)^2} \mathrm{~d} x=\int_0^{\pi / 4}|\cos x-\sin x| \mathrm{d} x \\ & =\int_0^{\pi / 4}(\cos x-\sin x) \mathrm{d} x=[\sin x+\cos x]_0^{\pi / 4} \\ & =\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-(0+1)=-1\end{aligned}$
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