Answer
$$\frac{\pi }{2}$$
Work Step by Step
$$\eqalign{
& \int_{1/2}^1 {\frac{{6dt}}{{\sqrt {3 + 4t - 4{t^2}} }}} \cr
& {\text{complete the square for }}3 + 4t - 4{t^2} \cr
& 3 + 4t - 4{t^2} = 3 - \left( {4{t^2} - 4t + 1} \right) + 1 \cr
& = 4 - \left( {4{t^2} - 4t + 1} \right) \cr
& = 4 - {\left( {2t - 1} \right)^2} \cr
& {\text{then}} \cr
& \int_{1/2}^1 {\frac{{6dt}}{{\sqrt {3 + 4t - 4{t^2}} }}} = \int_{1/2}^1 {\frac{{6dt}}{{\sqrt {4 - {{\left( {2t - 1} \right)}^2}} }}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = 2t - 1,{\text{ so that }}du = 2dt \cr
& {\text{the new limits on }}t{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}t = 1,{\text{ then }}u = 2\left( 1 \right) - 1 = 1 \cr
& \,\,\,\,\,\,{\text{If }}t = 1/2,{\text{ then }}u = 2\left( {1/2} \right) - 1 = 0 \cr
& {\text{then}} \cr
& \int_{1/2}^1 {\frac{{6dt}}{{\sqrt {4 - {{\left( {2t - 1} \right)}^2}} }}} = \int_0^1 {\frac{{6\left( {du/2} \right)}}{{\sqrt {4 - {u^2}} }}} \cr
& = 3\int_0^1 {\frac{{du}}{{\sqrt {4 - {u^2}} }}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 2 \cr
& = 3\left( {{{\sin }^{ - 1}}\left( {\frac{u}{2}} \right)} \right)_0^1 \cr
& = 3{\sin ^{ - 1}}\left( {\frac{1}{2}} \right) - 3{\sin ^{ - 1}}\left( {\frac{0}{2}} \right) \cr
& = 3\left( {\frac{\pi }{6}} \right) \cr
& = \frac{\pi }{2} \cr} $$
