## Calculus: Early Transcendentals (2nd Edition)

$= \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\,\left( {\frac{{\tan \,\left( {\frac{x}{2}} \right)}}{{\sqrt 3 }}} \right) + C$
$\begin{gathered} \int_{}^{} {\frac{{dx}}{{2 + \cos x}}} \hfill \\ \hfill \\ set\,\,x = 2{\tan ^{ - 1}}\,\left( u \right)\,\,\,\,then\,\,\,dx = 2\frac{{du}}{{1 + {u^2}}} \hfill \\ and\,\,\cos x = \frac{{1 - {u^2}}}{{1 + {u^2}}} \hfill \\ \hfill \\ substitute\,\,for\,dx{\text{ and }}\cos x \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{2 + \cos x}}} = \int_{}^{} {\frac{{\frac{{2du}}{{1 + {u^2}}}}}{{2 + \frac{{1 - {u^2}}}{{1 + {u^2}}}}}} = \int_{}^{} {\frac{2}{{{u^2} + 3}}du} \hfill \\ \hfill \\ integrating \hfill \\ \hfill \\ = \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\,\left( {\frac{u}{{\sqrt 3 }}} \right) + C \hfill \\ \hfill \\ substitute\,\,for\,\,\,u \hfill \\ \hfill \\ = \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\,\left( {\frac{{\tan \,\left( {\frac{x}{2}} \right)}}{{\sqrt 3 }}} \right) + C \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\,\left( {\frac{{\tan \,\left( {\frac{x}{2}} \right)}}{{\sqrt 3 }}} \right) + C \hfill \\ \hfill \\ \end{gathered}$