$\int {\frac{{dt}}{{\sqrt {3t - 2{t^2}} }}}$

Let $I = \int {\frac{{dt}}{{\sqrt {3t - 2{t^2}} }}} = \frac{1}{{\sqrt 2 }}\int {\frac{{dt}}{{\sqrt { - \left( {{t^2} - \frac{3}{2}t} \right)} }}}$

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

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

$= \frac{1}{{\sqrt 2 \int {\frac{{dt}}{{\sqrt {{{\left( {\frac{3}{4}} \right)}^2} - {{\left( {t - \frac{3}{4}} \right)}^2}} }}} }}$

$= \frac{1}{{\sqrt 2 }}{\sin ^{ - 1}}\left( {\frac{{t - \frac{3}{4}}}{{\frac{3}{4}}}} \right) + C = \frac{1}{{\sqrt 2 }}{\sin ^{ - 1}}\left( {\frac{{4t - 3}}{3}} \right) + C$