$\int_1^2 {\frac{{dx}}{{\sqrt {(x - 1)(2 - x)} }}}$

$I = \int_1^2 {\frac{{dx}}{{\sqrt {(x - 1)(2 - x)} }}} = \int_1^2 {\frac{{dx}}{{\sqrt {2x - {x^2} - 2 + x} }}}$

$= \int_1^2 {\frac{{dx}}{{\sqrt { - \left( {{x^2} - 3x + 2} \right)} }}}$

$= \int_1^2 {\frac{{dx}}{{\sqrt { - \left[ {{x^2} - 2 \cdot \frac{3}{2}x + {{\left( {\frac{3}{2}} \right)}^2} + 2 - \frac{9}{4}} \right]} }}}$

$= \int_1^2 {\frac{{dx}}{{\sqrt { - \left\{ {{{\left( {x - \frac{3}{2}} \right)}^2} - {{\left( {\frac{1}{2}} \right)}^2}} \right\}} }}}$

$= \int_1^2 {\frac{{dx}}{{\sqrt {{{\left( {\frac{1}{2}} \right)}^2} - {{\left( {x - \frac{3}{2}} \right)}^2}} }}} = \left[ {{{\sin }^{ - 1}}\left( {\frac{{x - \frac{3}{2}}}{{\frac{1}{2}}}} \right)} \right]_1^2$

$= \left[ {{{\sin }^{ - 1}}(2x - 3)} \right]_1^2 = {\sin ^{ - 1}}1 - {\sin ^{ - 1}}( - 1)$

$= \frac{\pi }{2} + \frac{\pi }{2}$

$= \pi$ and $\left. {\sin ( - \theta ) = - \sin \theta } \right]$