Answer
$$\frac{\pi }{6}$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\frac{{dx}}{{\sqrt {4x - {x^2}} }}} \cr
& {\text{completing the square for }}4x - {x^2} \cr
& 4x - {x^2} = - \left( {{x^2} - 4x + 4} \right) + 4 \cr
& 4x - {x^2} = 4 - {\left( {x - 2} \right)^2} \cr
& \int_1^2 {\frac{{dx}}{{\sqrt {4x - {x^2}} }}} = \int_1^2 {\frac{{dx}}{{\sqrt {4 - {{\left( {x - 2} \right)}^2}} }}} \cr
& \cr
& {\text{Write in terms of }}\theta \cr
& {\text{substitute }}x - 2 = 2\sin \theta ,{\text{ }}\,\,\,dx = 2\cos \theta d\theta \cr
& \,\,\,\,\theta = {\sin ^{ - 1}}\left( {\frac{{x - 2}}{2}} \right) \cr
& \,\,\,\,x = 2 \to \theta = 0 \cr
& \,\,\,\,x = 1 \to \theta = - \pi /6 \cr
& \int_1^2 {\frac{{dx}}{{\sqrt {4 - {{\left( {x - 2} \right)}^2}} }}} = \int_{ - \pi /6}^0 {\frac{{2\cos \theta d\theta }}{{\sqrt {4 - 4{{\sin }^2}\theta } }}} \cr
& = \int_{ - \pi /6}^0 {\frac{{2\cos \theta d\theta }}{{2\sqrt {1 - {{\sin }^2}\theta } }}} \cr
& = \int_{ - \pi /6}^0 {\frac{{2\cos \theta d\theta }}{{2\sqrt {{{\cos }^2}\theta } }}} \cr
& = \int_{ - \pi /6}^0 {d\theta } \cr
& = \left( \theta \right)_{ - \pi /6}^0 \cr
& \cr
& {\text{Evaluate the limits}} \cr
& = 0 - \left( { - \frac{\pi }{6}} \right) \cr
& = \frac{\pi }{6} \cr} $$
