Answer
$\frac{1}{2}{\csc ^4}\left( {\frac{x}{2}} \right) - \frac{1}{3}{\csc ^6}\left( {\frac{x}{2}} \right) + C$
Work Step by Step
$$\eqalign{
& \int {{{\cot }^3}\frac{x}{2}{{\csc }^4}\frac{x}{2}} dx \cr
& {\text{Split the integrand}}{\text{, use }}{a^{m + n}} = {a^m}{a^n} \cr
& \int {{{\cot }^2}\frac{x}{2}{{\csc }^3}\frac{x}{2}} \left( {\cot \frac{x}{2}\csc \frac{x}{2}} \right)dx \cr
& {\text{Use the pythagorean identity }}{\cot ^2}\theta = {\csc ^2}\theta - 1 \cr
& \int {\left( {{{\csc }^2}\frac{x}{2} - 1} \right){{\csc }^3}\frac{x}{2}} \left( {\cot \frac{x}{2}\csc \frac{x}{2}} \right)dx \cr
& \int {\left( {{{\csc }^5}\frac{x}{2} - {{\csc }^3}\frac{x}{2}} \right)} \left( {\cot \frac{x}{2}\csc \frac{x}{2}} \right)dx \cr
& {\text{Integrate by substitution}}{\text{, let }}u = \csc \frac{x}{2},{\text{ }} \cr
& du = - \left( {\csc \frac{x}{2}\cot \frac{x}{2}} \right)\left( {\frac{1}{2}} \right)dx,{\text{ }} - 2du = \csc \frac{x}{2}\cot \frac{x}{2}dx \cr
& {\text{Substituting}} \cr
& \int {\overbrace {\left( {{{\csc }^5}\frac{x}{2} - {{\csc }^3}\frac{x}{2}} \right)}^{\left( {{u^5} - {u^3}} \right)}} \overbrace {\left( {\cot \frac{x}{2}\csc \frac{x}{2}} \right)dx}^{ - 2du} = \int {\left( {{u^5} - {u^3}} \right)} \left( { - 2} \right)du \cr
& = 2\int {\left( {{u^3} - {u^5}} \right)} du \cr
& {\text{Integrate}} \cr
& = 2\left( {\frac{{{u^4}}}{4} - \frac{{{u^6}}}{6}} \right) + C \cr
& = \frac{1}{2}{u^4} - \frac{1}{3}{u^6} + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}\csc \frac{x}{2}{\text{ for }}u \cr
& = \frac{1}{2}{\csc ^4}\left( {\frac{x}{2}} \right) - \frac{1}{3}{\csc ^6}\left( {\frac{x}{2}} \right) + C \cr} $$
