## Thomas' Calculus 13th Edition

$$\pi$$
\eqalign{ & \int_{ - 3/4}^{3/4} {\frac{{6dx}}{{\sqrt {9 - 4{x^2}} }}} \cr & = \int_{ - 3/4}^{3/4} {\frac{{6dx}}{{\sqrt {{3^2} - {{\left( {2x} \right)}^2}} }}} \cr & {\text{Use the substitution method}}{\text{:}} \cr & u = 2x,{\text{ so that }}du = 2dx \cr & {\text{The new limits on }}t{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}x = 3/4,{\text{ then }}u = 2\left( {3/4} \right) = 3/2 \cr & \,\,\,\,\,\,{\text{If }}x = - 3/4,{\text{ then }}u = 2\left( { - 3/4} \right) = - 3/2 \cr & {\text{then}} \cr & \int_{ - 3/4}^{3/4} {\frac{{6dx}}{{\sqrt {{3^2} - {{\left( {2x} \right)}^2}} }}} = \int_{ - 3/2}^{3/2} {\frac{{6\left( {du/2} \right)}}{{\sqrt {{3^2} - {u^2}} }}} \cr & = 3\int_{ - 3/2}^{3/2} {\frac{{du}}{{\sqrt {{3^2} - {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 = 3 \cr & = 3\left( {{{\sin }^{ - 1}}\left( {\frac{u}{3}} \right)} \right)_{ - 3/2}^{3/2} \cr & = 3\left( {{{\sin }^{ - 1}}\left( {\frac{{3/2}}{3}} \right) - {{\sin }^{ - 1}}\left( {\frac{{ - 3/2}}{3}} \right)} \right) \cr & = 3\left( {\frac{\pi }{6} + \frac{\pi }{6}} \right) \cr & = \pi \cr}