$\begin{array}{l}
 \int {\frac{x}{{{{\sin }^2}x}}dx}  = \left[ {\begin{array}{*{20}{c}}
   {u = x} & {dv = \int {\frac{{dx}}{{{{\sin }^2}x}}} }  \\
   {du = dx} & {v =  - {\mathop{\rm ctg}\nolimits} x}  \\
\end{array}} \right] =  - x{\mathop{\rm ctg}\nolimits} x - \int {{\mathop{\rm ctg}\nolimits} xdx}  =  - x{\mathop{\rm ctg}\nolimits} x + \int {\frac{{\cos x}}{{\sin x}}dx}  = \left| \begin{array}{l}
 \sin x = t \\ 
 \cos xdx = dt \\ 
 \end{array} \right| =  \\ 
  =  - x{\mathop{\rm ctg}\nolimits} x + \int {\frac{{dt}}{t}}  =  - x{\mathop{\rm ctg}\nolimits} x + \ln (\sin x) = \ln (\sin x) - x{\mathop{\rm ctg}\nolimits} x + C \\ 
 \end{array}$