## Thomas' Calculus 13th Edition

$$- \frac{\pi }{{12\sqrt 3 }}$$
\eqalign{ & \int_{ - 2/\sqrt 5 }^{ - \sqrt 6 /\sqrt 5 } {\frac{{dy}}{{\left| y \right|\sqrt {5{y^2} - 3} }}} \cr & = \int_{ - 2/\sqrt 5 }^{ - \sqrt 6 /\sqrt 5 } {\frac{{dy}}{{\left| y \right|\sqrt {{{\left( {\sqrt 5 y} \right)}^2} - 3} }}} \cr & {\text{Use the substitution method:}} \cr & u = \sqrt 5 y,\,\,\,\,\,\,du = \sqrt 5 dy,\,\,\,\,dy = \frac{{du}}{{\sqrt 5 }} \cr & {\text{The new limits on }}u{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}y = - \sqrt 6 /\sqrt 5,{\text{ then }}u = \sqrt 5 \left( { - \sqrt 6 /\sqrt 5 } \right) = - \sqrt 6 \cr & \,\,\,\,\,\,{\text{If }}y = - 2/\sqrt 5,{\text{ then }}u = \sqrt 5 \left( { - 2/\sqrt 5 } \right) = - 2 \cr & {\text{Then}} \cr & \int_{ - 2/\sqrt 5 }^{ - \sqrt 6 /\sqrt 5 } {\frac{{dy}}{{\left| y \right|\sqrt {{{\left( {\sqrt 5 y} \right)}^2} - 3} }}} = \int_{ - 2}^{ - \sqrt 6 } {\frac{{du/\sqrt 5 }}{{\left| y \right|\sqrt {{u^2} - 3} }}} \cr & = \int_{ - 2}^{ - \sqrt 6 } {\frac{{du}}{{\left| {\sqrt 5 y} \right|\sqrt {{u^2} - 3} }}} \cr & = \int_{ - 2}^{ - \sqrt 6 } {\frac{{dy}}{{u\sqrt {{u^2} - 3} }}} \cr & {\text{integrate using }}\int {\frac{{du}}{{\left| u \right|\sqrt {{u^2} - {a^2}} }} = \frac{1}{a}{{\sec }^{ - 1}}\left| {\frac{u}{a}} \right| + C} \cr & = \left( {\frac{1}{{\sqrt 3 }}{{\sec }^{ - 1}}\left| {\frac{u}{{\sqrt 3 }}} \right|} \right)_{ - 2}^{ - \sqrt 6 } \cr & {\text{use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr & = \frac{1}{{\sqrt 3 }}\left( {{{\sec }^{ - 1}}\left| {\frac{{ - \sqrt 6 }}{{\sqrt 3 }}} \right| - {{\sec }^{ - 1}}\left| {\frac{{ - 2}}{{\sqrt 3 }}} \right|} \right) \cr & {\text{simplifying, we get:}} \cr & = \frac{1}{{\sqrt 3 }}\left( {\frac{3}{4}\pi - \frac{5}{6}\pi } \right) \cr & = - \frac{\pi }{{12\sqrt 3 }} \cr}