Question

${\int }_0^{\pi /2}\sqrt{1-sin2x}dx\left(a\right)2\sqrt{2}\left(b\right)2(\sqrt{2}+1)\left(c\right)2\left(d\right)2(\sqrt{2}-1)$

Answer 1

$I={\int }_0^{\pi /2}\sqrt{1-sin2x}dx={\int }_0^{\pi /4}\sqrt{(cosx-sinx{)}^2}dx+{\int }_{\pi /4}^{\pi /2}\sqrt{(sinx-cosx{)}^2}dx=[sinx+cosx{]}_0^{\pi /4}+[-cosx-sinx{]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-1+(-0-1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})=2\sqrt{2}-2=2\left(\sqrt{2-1}\right)$