Answer
$$\frac{\pi }{6}$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\frac{1}{{\sqrt x \sqrt {4 - x} }}} dx \cr
& or \cr
& = \int_1^2 {\frac{1}{{\sqrt x \sqrt {{2^2} - {{\left( {\sqrt x } \right)}^2}} }}} dx \cr
& {\text{substitute }}u = \sqrt x ,{\text{ }}du = \frac{1}{{2\sqrt x }}dx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = 2,{\text{ }}u = \sqrt 2 \cr
& {\text{if }}x = 1,{\text{ }}u = 1 \cr
& {\text{so}} \cr
& = \int_1^2 {\frac{1}{{\sqrt x \sqrt {{2^2} - {{\left( {\sqrt x } \right)}^2}} }}} dx = \int_1^{\sqrt 2 } {\frac{2}{{\sqrt {{{\left( 2 \right)}^2} - {u^2}} }}} du \cr
& {\text{find the antiderivative using the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\frac{u}{a} + C,{\text{ }}} \left( {{\text{see page 468}}} \right) \cr
& = \left. {2\left( {si{n^{ - 1}}\frac{u}{2}} \right)} \right|_1^{\sqrt 2 } \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = 2\left( {{{\sin }^{ - 1}}\left( {\frac{{\sqrt 2 }}{2}} \right) - {{\sin }^{ - 1}}\left( {\frac{1}{2}} \right)} \right) \cr
& {\text{simplify}} \cr
& = 2\left( {\frac{\pi }{4} - \frac{\pi }{6}} \right) \cr
& = \frac{\pi }{6} \cr} $$
