Answer
$$ - \frac{{{{\csc }^3}t\cot t}}{4} - \frac{{3\csc t\cot t}}{8} + \frac{3}{8}\ln \left| {\csc t - \cot t} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {{{\csc }^5}t} dt \cr
& {\text{Use the Table of Integrals on the Reference Pages }} \cr
& {\text{*}}\int {{{\csc }^n}u} du = - \frac{{{{\csc }^{n - 2}}u\cot u}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int {{{\csc }^{n - 2}}} udu,{\text{ }}n \ne 1 \cr
& {\text{Therefore,}} \cr
& \int {{{\csc }^5}t} dt = - \frac{{{{\csc }^{5 - 2}}t\cot t}}{{5 - 1}} + \frac{{5 - 2}}{{5 - 1}}\int {{{\csc }^{5 - 2}}} tdt \cr
& \int {{{\csc }^5}t} dt = - \frac{{{{\csc }^3}t\cot t}}{4} + \frac{3}{4}\int {{{\csc }^3}} tdt \cr
& {\text{For }}\int {{{\csc }^3}} tdt{\text{ use the Table of Integrals}} \cr
& = - \frac{{{{\csc }^3}t\cot t}}{4} + \frac{3}{4}\left( { - \frac{{{{\csc }^{3 - 2}}t\cot t}}{{3 - 1}} + \frac{{3 - 2}}{{3 - 1}}\int {{{\csc }^{3 - 2}}} tdt} \right) \cr
& = - \frac{{{{\csc }^3}t\cot t}}{4} + \frac{3}{4}\left( { - \frac{{\csc t\cot t}}{2} + \frac{1}{2}\int {\csc } tdt} \right) \cr
& {\text{Simplifying}} \cr
& = - \frac{{{{\csc }^3}t\cot t}}{4} - \frac{{3\csc t\cot t}}{8} + \frac{3}{8}\int {\csc } tdt \cr
& {\text{Where }}\int {\csc u} du = \ln \left| {\csc u - \cot u} \right| + C \cr
& = - \frac{{{{\csc }^3}t\cot t}}{4} - \frac{{3\csc t\cot t}}{8} + \frac{3}{8}\ln \left| {\csc t - \cot t} \right| + C \cr} $$
