Answer
\[ = \frac{1}{2}\ln \left| {\frac{{\sin x}}{{\sin x + 2}}} \right| + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{\cos x}}{{{{\sin }^2}x + 2\sin x}}\,dx} \hfill \\
\hfill \\
set\,\,u = \sin x\,\,\,\,then\,\,\,\,du = \cos xdx \hfill \\
\hfill \\
\int_{}^{} {\frac{{\cos x}}{{{{\sin }^2}x + 2\sin x}}\,dx} = \int_{}^{} {\frac{{du}}{{{u^2} + 2u}}} = \int_{}^{} {\frac{{du}}{{u\,\left( {u + 2} \right)}}} \hfill \\
\hfill \\
use\,\,the\,\,formula\,\,\int_{}^{} {\frac{1}{{u\,\left( {a + b} \right)}}du = \frac{1}{a}\ln \left| {\frac{u}{{a + bu}}} \right| + C} \hfill \\
then \hfill \\
\hfill \\
= \frac{1}{2}\ln \left| {\frac{u}{{u + 2}}} \right| + C \hfill \\
\hfill \\
substitute\,\,back\,\,u = \sin x \hfill \\
\hfill \\
= \frac{1}{2}\ln \left| {\frac{{\sin x}}{{\sin x + 2}}} \right| + C \hfill \\
\hfill \\
\end{gathered} \]
