Answer
$$\sqrt 2 $$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {\sqrt {1 - \cos 2x} dx} \cr
& {\text{Use the identity }}\cos 2x = 1 - 2{\sin ^2}x \cr
& \int_0^{\pi /2} {\sqrt {1 - \cos 2x} dx} = \int_0^{\pi /2} {\sqrt {1 - 1 + 2{{\sin }^2}x} dx} \cr
& = \int_0^{\pi /2} {\sqrt {2{{\sin }^2}x} dx} \cr
& = \sqrt 2 \int_0^{\pi /2} {\sin xdx} \cr
& {\text{Integrate}} \cr
& = \sqrt 2 \left[ { - \cos x} \right]_0^{\pi /2} \cr
& = \sqrt 2 \left[ { - \cos \left( {\frac{\pi }{2}} \right) + \cos \left( 0 \right)} \right] \cr
& = \sqrt 2 \cr} $$
