Question

${\int }_0^{\frac{\pi }{2}}\sqrt{1-\sin 2x}dx$ is equal to

Answer 1

Given : ${\int }_0^{\frac{\pi }{2}}\sqrt{1-\sin 2x}dx$

$I={\int }_0^{\frac{\pi }{2}}\sqrt{1-\sin 2x}dx={\int }_0^{\frac{\pi }{2}}\sqrt{{|\sin x-\cos x|}^2}dxI={\int }_0^{\frac{\pi }{2}}|\sin x-\cos x|dx={\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}|\sin x-\cos x|-{\int }_0^{\frac{\pi }{4}}|\sin x-\cos x|I={[-\cos x-\sin x]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}-{|-\cos x-\sin x|}_0^{\frac{\pi }{4}}I=-0-1+\sqrt{2}-(-\sqrt{2}+1)=2\sqrt{2}-2$

Hence the correct answer is $2\sqrt{2}-2$