Answer
$ - \sqrt 2 \ln \left| {\csc \sqrt {2x} + \cot \sqrt {2x} } \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{\csc \sqrt {2x} }}{{\sqrt x }}} dx \cr
& = \sqrt 2 \int {\frac{{\csc \sqrt {2x} }}{{\sqrt {2x} }}} dx \cr
& = \sqrt 2 \int {\csc \sqrt {2x} \left( {\frac{1}{{\sqrt {2x} }}} \right)} dx \cr
& {\text{Let }}u = \sqrt {2x} ,{\text{ }}du = \frac{1}{{\sqrt {2x} }}dx,{\text{ }}dx = \sqrt {2x} du \cr
& {\text{Substituting}} \cr
& \sqrt 2 \int {\csc \sqrt {2x} \left( {\frac{1}{{\sqrt {2x} }}} \right)} dx \cr
& = \sqrt 2 \int {\csc udu} \cr
& {\text{Where }}\int {\csc u} du = - \ln \left| {\csc u + \cot u} \right| + C,{\text{ then}} \cr
& = \sqrt 2 \left( { - \ln \left| {\csc u + \cot u} \right|} \right) + C \cr
& = - \sqrt 2 \ln \left| {\csc u + \cot u} \right| + C \cr
& {\text{Write in terms of }}x,{\text{ susbstitute }}\sqrt {2x} {\text{ for }}u \cr
& = - \sqrt 2 \ln \left| {\csc \sqrt {2x} + \cot \sqrt {2x} } \right| + C \cr} $$
