What is the integration of \[ \int {\frac{{dt}}{{\sqrt {3t – 2{t^2}} }}} \] ~~~~~[NCERT Exemp. Q. 15,Page 164 ]

Let \[ I = \int {\frac{{dt}}{{\sqrt {3t – 2{t^2}} }}} = \frac{1}{{\sqrt 2 }}\int {\frac{{dt}}{{\sqrt { – \left( {{t^2} – \frac{3}{2}t} \right)} }}} \]


\[ = \frac{1}{{\sqrt 2 }}\int {\frac{{dt}}{{\sqrt { – \left[ {\left( {{t^2} – 2 \cdot \frac{1}{2} \cdot \frac{3}{2}t} \right) + {{\left( {\frac{3}{4}} \right)}^2} – {{\left( {\frac{3}{4}} \right)}^2}} \right]} }}} \]


\[ = \frac{1}{{\sqrt 2 \int {\frac{{dt}}{{\sqrt { – \left[ {{{\left( {t – \frac{3}{4}} \right)}^2} – {{\left( {\frac{3}{4}} \right)}^2}} \right]} }}} }}\]

\[ = \frac{1}{{\sqrt 2 \int {\frac{{dt}}{{\sqrt {{{\left( {\frac{3}{4}} \right)}^2} – {{\left( {t – \frac{3}{4}} \right)}^2}} }}} }}\]


\[ = \frac{1}{{\sqrt 2 }}{\sin ^{ – 1}}\left( {\frac{{t – \frac{3}{4}}}{{\frac{3}{4}}}} \right) + C = \frac{1}{{\sqrt 2 }}{\sin ^{ – 1}}\left( {\frac{{4t – 3}}{3}} \right) + C\]

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