Answer
\[ = \frac{1}{{2{{\cot }^2}x}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{{\csc }^2}x}}{{{{\cot }^3}x}}} \,\,dx \hfill \\
\hfill \\
u = \cot x\,\,\,\,then\,\,\,\,du = - {\csc ^2}xdx \hfill \\
\hfill \\
apply\,\,the\,\,\,substitution \hfill \\
\hfill \\
= \int_{}^{} { - \frac{{du}}{{{u^3}}}} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{1}{{2{u^2}}} + C \hfill \\
\hfill \\
replace\,\,u\,\,with\,\,\,u = \cot x \hfill \\
\hfill \\
= \frac{1}{{2{{\cot }^2}x}} + C \hfill \\
\hfill \\
\end{gathered} \]
