Answer
$ - \frac{{{{\cos }^4}x}}{4} + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\cos }^3}x}}{{\csc x}}} dx \cr
& {\text{Use the reciprocal identity }}\csc x = \frac{1}{{\sin x}} \cr
& \int {\frac{{{{\cos }^3}x}}{{\csc x}}} dx = \int {\frac{{{{\cos }^3}x}}{{\left( {1/\sin x} \right)}}} dx \cr
& = \int {{{\cos }^3}x\sin x} dx \cr
& {\text{Use the substitution method}} \cr
& {\text{Let }}u = \cos x,{\text{ }}du = - \sin xdx \cr
& {\text{Substituting}} \cr
& \int {{{\cos }^3}x\sin x} dx = \int {{u^3}} \left( { - 1} \right)du \cr
& = - \int {{u^3}} du \cr
& {\text{Integrate apply the power rule }}\int {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}}{\text{ + C }} \cr
& = - \frac{{{u^4}}}{4} + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}\cos x{\text{ for }}u \cr
& = - \frac{{{{\cos }^4}x}}{4} + C \cr} $$
