$\int_0^{\pi/2} \sqrt{1 - \sin 2x}\,dx$ is equal to:

$\sqrt{1 - \sin 2x}$ can be simplified using $\sin 2x = 2\sin x \cos x$:

$1 - \sin 2x = 1 - 2\sin x \cos x$

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Expanding $(\sin x - \cos x)^2$:

$(\sin x - \cos x)^2 = \sin^2 x - 2\sin x \cos x + \cos^2 x$

$= 1 - 2\sin x \cos x$

Therefore:

$\sqrt{1 - \sin 2x} = \sqrt{(\sin x - \cos x)^2}$

$= |\sin x - \cos x|$

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For $x \in [0, \pi/2]$, the sign of $\sin x - \cos x$ changes at $x = \pi/4$:

• At $x = 0$: $\sin 0 = 0 < \cos 0 = 1$
• At $x = \pi/4$: $\sin(\pi/4) = \cos(\pi/4)$
• At $x = \pi/2$: $\sin(\pi/2) = 1 > \cos(\pi/2) = 0$

So:

$|\sin x - \cos x| = \begin{cases} \cos x - \sin x & \text{for } 0 \leq x \leq \pi/4 \\ \sin x - \cos x & \text{for } \pi/4 \leq x \leq \pi/2 \end{cases}$

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The integral splits at $x = \pi/4$:

$\int_0^{\pi/2} \sqrt{1 - \sin 2x}\,dx = \int_0^{\pi/4} (\cos x - \sin x)\,dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x)\,dx$

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For the first integral:

$\int_0^{\pi/4} (\cos x - \sin x)\,dx = [\sin x + \cos x]_0^{\pi/4}$

$= \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) - (0 + 1)$

$= \sqrt{2} - 1$

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For the second integral:

$\int_{\pi/4}^{\pi/2} (\sin x - \cos x)\,dx = [-\cos x - \sin x]_{\pi/4}^{\pi/2}$

$= (0 - 1) - \left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right)$

$= -1 + \sqrt{2}$

$= \sqrt{2} - 1$

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The total value:

$(\sqrt{2} - 1) + (\sqrt{2} - 1) = 2(\sqrt{2} - 1)$

Therefore, the answer is $2(\sqrt{2} - 1)$.