$\int\limits_{0}^{\pi/2} \sqrt{1 - \sin 2x} \, dx$ is equal to

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

The correct answer is Option (4) → $2(\sqrt{2} - 1)$

Let $I = \int\limits_{0}^{\pi/2} \sqrt{1 - \sin 2x} \, dx$

$= \int\limits_{0}^{\pi/2} \sqrt{\sin^2 x + \cos^2 x - 2 \sin x \cos x} \, dx$

$[∵\sin^2 x + \cos^2 x = 1, \sin 2x = 2 \sin x \cdot \cos x]$

$= \int\limits_{0}^{\pi/4} \sqrt{(\cos x - \sin x)^2} \, dx + \int\limits_{\pi/4}^{\pi/2} \sqrt{(\sin x - \cos x)^2} \, dx$

$\left[ ∵\text{when } 0 < x < \frac{\pi}{4}, \cos x > \sin x \text{ and when } \frac{\pi}{4} < x < \frac{\pi}{2}, \sin x > \cos x \right]$

$= [\sin x + \cos x]_0^{\pi/4} + [-\cos x - \sin x]_{\pi/4}^{\pi/2}$

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

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