Solution:
Let $\text{I}=\int\limits^{\frac{\pi}{2}}_0\sqrt{1-\sin2\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\sqrt{(\cos\text{x}-\sin\text{x})^2}\text{ dx}+\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{4}}\sqrt{(\sin\text{x}-\cos\text{x})^2}\text{ dx}$
$=\Big[\sin\text{x}+\cos\text{x}\Big]^{\frac{\pi}{4}}_0+\Big[-\cos\text{x}-\sin\text{x}\Big]^{\frac{\pi}{2}}_{\frac{\pi}{4}}$
$=\frac{1}{\sqrt{2}}+=\frac{1}{\sqrt{2}}-0-1+\Big(-0-1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\Big)$
$=2\sqrt{2}-2=2(\sqrt{2}-1)$
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