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