Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_{\pi /2}^{3\pi /4} {\sqrt {1 - \sin 2x} } dx \cr
& {\text{Multiply the integrand by }}\sqrt {\frac{{1 + \sin 2x}}{{1 + \sin 2x}}} \cr
& = \int_{\pi /2}^{3\pi /4} {\sqrt {1 - \sin 2x} } \sqrt {\frac{{1 + \sin 2x}}{{1 + \sin 2x}}} dx \cr
& = \int_{\pi /2}^{3\pi /4} {\sqrt {\frac{{\left( {1 - \sin 2x} \right)\left( {1 + \sin 2x} \right)}}{{1 + \sin 2x}}} } dx \cr
& {\text{Recall that }}\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2} \cr
& = \int_{\pi /2}^{3\pi /4} {\sqrt {\frac{{1 - {{\sin }^2}2x}}{{1 + \sin 2x}}} } dx \cr
& {\text{use the indentity co}}{{\text{s}}^2}\theta = 1 - {\sin ^2}\theta \cr
& = \int_{\pi /2}^{3\pi /4} {\sqrt {\frac{{{{\cos }^2}2x}}{{1 + \sin 2x}}} } dx \cr
& {\text{Where }}\sqrt {{{\cos }^2}2x} = - \cos 2x,{\text{ for the interval }}\left( {\pi {\text{/2}}{\text{,3}}\pi {\text{/4}}} \right),{\text{ then}} \cr
& = \int_{\pi /2}^{3\pi /4} {\frac{{ - \cos 2x}}{{\sqrt {1 + \sin 2x} }}} dx \cr
& \cr
& {\text{Use the substitution method}} \cr
& {\text{Let }}u = 1 + \sin 2x,\,\,\,\,du = 2\cos 2xdx \cr
& \,\,{\text{Change the limits of integration}} \cr
& \,\,\,\,\,\,\,\,u = 1 + \sin 2x,\,\,\,x = 3\pi /4 \to u = 0 \cr
& \,\,\,\,\,\,\,\,u = 1 + \sin 2x,\,\,\,x = \pi /2 \to u = 1 \cr
& {\text{Write the integral in terms of }}u \cr
& \int_{\pi /2}^{3\pi /4} {\frac{{ - \cos 2x}}{{\sqrt {1 + \sin 2x} }}} dx = - \int_1^0 {\frac{{du}}{{2\sqrt u }}} \cr
& - \int_1^0 {\frac{{du}}{{2\sqrt u }}} = - \left( {\sqrt u } \right)_1^0 \cr
& = - \left( {\sqrt 0 - \sqrt 1 } \right) \cr
& = 1 \cr} $$
