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