Answer
$$\frac{{\sqrt 3 }}{3}\pi $$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^2 {\frac{{3dt}}{{4 + 3{t^2}}}} \cr
& = \int_{ - 2}^2 {\frac{{3dt}}{{{{\left( 2 \right)}^2} + {{\left( {\sqrt 3 t} \right)}^2}}}} \cr
& {\text{Use the substitution method}}{\text{:}} \cr
& u = \sqrt 3 t,{\text{ so that }}du = \sqrt 3 dt \cr
& {\text{The new limits on }}t{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}t = 2,{\text{ then }}u = \sqrt 3 \left( 2 \right) = 2\sqrt 3 \cr
& \,\,\,\,\,\,{\text{If }}t = - 2,{\text{ then }}u = \sqrt 3 \left( { - 2} \right) = - 2\sqrt 3 \cr
& {\text{then}} \cr
& \int_{ - 2}^2 {\frac{{3dt}}{{{{\left( 2 \right)}^2} + {{\left( {\sqrt 3 t} \right)}^2}}}} = \int_{ - 2\sqrt 3 }^{2\sqrt 3 } {\frac{{3\left( {du/\sqrt 3 } \right)}}{{{{\left( 2 \right)}^2} + {u^2}}}} \cr
& = \sqrt 3 \int_{ - 2\sqrt 3 }^{2\sqrt 3 } {\frac{{du}}{{{{\left( 2 \right)}^2} + {u^2}}}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 2 \cr
& = \sqrt 3 \left( {\frac{1}{2}ta{n^{ - 1}}\left( {\frac{u}{2}} \right)} \right)_{ - 2\sqrt 3 }^{2\sqrt 3 } \cr
& = \frac{{\sqrt 3 }}{2}\left( {ta{n^{ - 1}}\left( {\frac{{2\sqrt 3 }}{2}} \right) - ta{n^{ - 1}}\left( {\frac{{ - 2\sqrt 3 }}{2}} \right)} \right) \cr
& {\text{simplifying, we get:}} \cr
& = \frac{{\sqrt 3 }}{2}\left( {ta{n^{ - 1}}\left( {\sqrt 3 } \right) - ta{n^{ - 1}}\left( { - \sqrt 3 } \right)} \right) \cr
& = \frac{{\sqrt 3 }}{2}\left( {\frac{\pi }{3} + \frac{\pi }{3}} \right) \cr
& = \frac{{\sqrt 3 }}{2}\left( {\frac{{2\pi }}{3}} \right) \cr
& = \frac{{\sqrt 3 }}{3}\pi \cr} $$
