## Calculus: Early Transcendentals (2nd Edition)

$= \frac{\pi }{6}$
$\begin{gathered} \int_{\frac{2}{{5\sqrt 3 }}}^{\frac{2}{5}} {\frac{{dx}}{{x\sqrt {25{x^2} - 1} }}} \hfill \\ \hfill \\ rewrite \hfill \\ \hfill \\ \int_{\frac{2}{{5\sqrt 3 }}}^{\frac{2}{5}} {\frac{{5dx}}{{5x\sqrt {{{\left( {5x\,} \right)}^2} - 1} }}} \hfill \\ \hfill \\ set\,\,u = 5x\,\,\,\,then\,\,\,du = 5dx \hfill \\ \hfill \\ x = \frac{2}{{5\sqrt 3 }}\,\,\,\,\,implies\,\,\,u = \frac{2}{{\sqrt 3 }} \hfill \\ \hfill \\ x = \frac{2}{5}\,\,\,\,\,\,\,\,\,\,\,\,\,implies\,\,u = 2 \hfill \\ \hfill \\ apply\,\,the\,\,\,substitution \hfill \\ \hfill \\ \int_{\frac{2}{{\sqrt 3 }}}^2 {\frac{{du}}{{u\sqrt {{u^2} - 1} }}} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \,\,\left[ {\operatorname{arcsec} \,u} \right]_{\frac{2}{{\sqrt 3 }}}^2 \hfill \\ \hfill \\ Fundamental\,\,theorem \hfill \\ \hfill \\ = \operatorname{arcsec} \,\left( 2 \right) - \operatorname{arcsec} \,\left( {\frac{2}{{\sqrt 3 }}} \right) \hfill \\ \hfill \\ = \frac{\pi }{3} - \frac{\pi }{6} \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ = \frac{\pi }{6} \hfill \\ \end{gathered}$