Answer
$ - \frac{1}{4}\cot x{\csc ^3}x - \frac{3}{8}\cot x\csc x + \frac{3}{8}\ln \left| {\csc x - \cot x} \right| + C$
Work Step by Step
$$\eqalign{
& \int {{{\csc }^5}x} dx \cr
& {\text{Integrate by using a CAS }}\left( {{\text{See image below}}} \right){\text{ we obtain:}} \cr
& {\text{Integrating by tables using the entry 78}} \cr
& {\text{Entry 78:}}\int {{{\csc }^n}u} du = \frac{{ - 1}}{{n - 1}}\cot u{\csc ^{n - 2}}u + \frac{{n - 2}}{{n - 1}}\int {{{\csc }^{n - 2}}u} du \cr
& {\text{Let }}n = 5 \cr
& \int {{{\csc }^5}x} dx = \frac{{ - 1}}{{5 - 1}}\cot x{\csc ^{5 - 2}}x + \frac{{5 - 2}}{{5 - 1}}\int {{{\csc }^{5 - 2}}x} dx \cr
& \int {{{\csc }^5}x} dx = - \frac{1}{4}\cot x{\csc ^3}x + \frac{3}{4}\int {{{\csc }^3}x} dx \cr
& {\text{Use the entry 72}} \cr
& = - \frac{1}{4}\cot x{\csc ^3}x + \frac{3}{4}\left( {\frac{{ - 1}}{2}\cot x\csc x + \frac{1}{2}\ln \left| {\csc x - \cot x} \right|} \right) + C \cr
& {\text{Simplifying}} \cr
& = - \frac{1}{4}\cot x{\csc ^3}x - \frac{3}{8}\cot x\csc x + \frac{3}{8}\ln \left| {\csc x - \cot x} \right| + C \cr} $$
