What is the integration of \[ \int_1^2 {\frac{{dx}}{{\sqrt {(x – 1)(2 – x)} }}} \]~~~~~[NCERT Exemp. Q. 31,Page 165 ]

\[ I = \int_1^2 {\frac{{dx}}{{\sqrt {(x – 1)(2 – x)} }}} = \int_1^2 {\frac{{dx}}{{\sqrt {2x – {x^2} – 2 + x} }}} \]


\[ = \int_1^2 {\frac{{dx}}{{\sqrt { – \left( {{x^2} – 3x + 2} \right)} }}} \]


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


\[ = \int_1^2 {\frac{{dx}}{{\sqrt { – \left\{ {{{\left( {x – \frac{3}{2}} \right)}^2} – {{\left( {\frac{1}{2}} \right)}^2}} \right\}} }}} \]


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


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


\[ = \frac{\pi }{2} + \frac{\pi }{2}\]


\[ = \pi \] and \[ \left. {\sin ( – \theta ) = – \sin \theta } \right]\]


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