Answer
$$ - \frac{1}{2}\cos \left( {{x^2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{\csc \left( {{x^2}} \right)}}} dx \cr
& {\text{trigonometric identity sin}}\phi = \frac{1}{{\csc \phi }} \cr
& = \int {\sin \left( {{x^2}} \right)x} dx \cr
& {\text{substitute }}u = {x^2}{\text{ }} \cr
& du = 2xdx \cr
& \frac{1}{2}du = xdx \cr
& = \frac{1}{2}\int {\sin u} du \cr
& {\text{find the antiderivative}} \cr
& = \frac{1}{2}\left( { - \cos u} \right) + C \cr
& = - \frac{1}{2}\cos u + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = {x^2}{\text{ }} \cr
& = - \frac{1}{2}\cos \left( {{x^2}} \right) + C \cr} $$
