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


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

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

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

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

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

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

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

Buy Best Mathematics E-Books Visit : https://mathstudy.in/

Buy Mathematics Formula Book for Class XI,XII,JEE and other Engineering Competition Exam https://mathstudy.in/product/mathematics-formula-book/

Buy Mathematics Workbook for Class XII ( Fully Solved ) : https://mathstudy.in/product/work-book-class-xii-c-b-s-e-fully-solved/

Buy Mathematics Chapter Tests for Class XII ( Fully Solved) : https://mathstudy.in/product/mathematics-chapter-tests-class-xii-c-b-s-e/

Buy Objective Type Question Bank Class XII (Fully Solved ) : https://mathstudy.in/product/objective-type-question-bank-for-mathematics-class-xii-c-b-s-e/