Answer
$$ - \ln \left| {\csc x + \cot x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\csc x} dx \cr
& {\text{Multiply the integrand by }}\frac{{\csc x + \cot x}}{{\csc x + \cot x}} \cr
& = \int {\csc x} \left( {\frac{{\csc x + \cot x}}{{\csc x + \cot x}}} \right)dx \cr
& = \int {\frac{{{{\csc }^2}x + \csc x\cot x}}{{\csc x + \cot x}}} dx \cr
& \cr
& {\text{Let }}u = \csc x + \cot x,\,\,\,\,du = \left( { - \csc x\cot x - {{\csc }^2}x} \right)dx \cr
& {\text{Use the substitution}} \cr
& \int {\frac{{{{\csc }^2}x + \csc x\cot x}}{{\csc x + \cot x}}} dx = \int {\frac{{ - du}}{u}} \cr
& {\text{Integrate}} \cr
& {\text{ = }} - \ln \left| u \right| + C \cr
& {\text{substituting back }}u = \csc x + \cot x \cr
& {\text{ = }} - \ln \left| {\csc x + \cot x} \right| + C \cr} $$
