$\int\limits^{\pi /2}_{0}\sqrt{1-sin2x}\, \, dx$ is equal to :

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

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

$\int\limits^{\pi /2}_{0}\sqrt{1-\sin 2x}\, dx$

$∵\sqrt{1-\sin 2x}=\sqrt{\cos^2x+\sin^2x-2\sin x\cos x}=|\cos x-\sin x|$

$=\int\limits^{\pi /2}_{0}|\cos x-\sin x|dx$

$=\int\limits^{\pi /4}_{0}\cos x-\sin x\,dx+\int\limits^{\pi /2}_{\pi /4}\sin x-\cos x\,dx$

$=\left[\sin x+\cos x\right]^{\pi /4}_{0}+\left[\sin x+\cos x\right]^{\pi /2}_{\pi /4}$

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