## Calculus: Early Transcendentals (2nd Edition)

$= \frac{5}{9}\pi - \frac{{\sqrt 3 }}{3}$
$\begin{gathered} \int_{\frac{2}{{\sqrt 3 }}}^2 {z{{\sec }^{ - 1}}zdz} \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ u = {\sec ^{ - 1}}z\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,du = \frac{{du}}{{z\sqrt {{z^2} - 1} }} \hfill \\ dv = dz\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,v = \frac{{{z^2}}}{2} \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ = \frac{{{z^2}{{\sec }^{ - 1}}z}}{2} - \int_{}^{} {\,\left( {\frac{{{z^2}}}{2}} \right)\,\left( {\frac{{dz}}{{z\sqrt {{z^2} - 1} }}} \right)} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ = \frac{{{z^2}{{\sec }^{ - 1}}z}}{2} - \frac{1}{2}\int_{}^{} {\,\frac{{zdz}}{{\sqrt {{z^2} - 1} }}} \hfill \\ = \frac{{{z^2}{{\sec }^{ - 1}}z}}{2} - \frac{1}{4}\int_{}^{} {\,{{\left( {{z^2} - 1} \right)}^{ - 1/2}}\left( {2z} \right)dz} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \frac{{{z^2}{{\sec }^{ - 1}}z}}{2} - \frac{1}{4}\,\left( {\frac{{\sqrt {{z^2} - 1} }}{{\frac{1}{2}}}} \right) + C \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{{{z^2}{{\sec }^{ - 1}}z}}{2} - \frac{1}{2}\sqrt {{z^2} - 1} + C \hfill \\ \hfill \\ use\,\,the\,\,ftc \hfill \\ \hfill \\ = \,\,\left[ {\frac{{{z^2}{{\sec }^{ - 1}}z}}{2} - \frac{1}{2}\sqrt {{z^2} - 1} } \right]_{\frac{2}{{\sqrt 3 }}}^2 \hfill \\ \hfill \\ Evaluate\,\,\,the\,\,limits \hfill \\ \hfill \\ = \,\,\left[ {\frac{{\,{{\left( 2 \right)}^2}{{\sec }^{ - 1}}\,\left( 2 \right)}}{2} - \frac{1}{2}\sqrt 3 } \right] - \,\,\left[ {\frac{{\frac{4}{3}{{\sec }^{ - 1}}\,\left( {\frac{2}{{\sqrt 3 }}} \right)}}{2} - \frac{1}{2}\,\left( {\frac{{\sqrt 3 }}{3}} \right)} \right] \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{{2\pi }}{3} - \frac{{\sqrt 3 }}{2} - \frac{\pi }{9} + \frac{{\sqrt 3 }}{6} \hfill \\ \hfill \\ solution \hfill \\ \hfill \\ = \frac{5}{9}\pi - \frac{{\sqrt 3 }}{3} \hfill \\ \end{gathered}$