Answer
$$ - 2\csc \sqrt \theta + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\cos \sqrt \theta }}{{\sqrt \theta {{\sin }^2}\sqrt \theta }}} d\theta \cr
& = \int {\frac{{\cos \sqrt \theta }}{{\sqrt \theta {{\left( {\sin \sqrt \theta } \right)}^2}}}} d\theta \cr
& \cr
& {\text{Integrate by the substitution method}} \cr
& u = \sin \sqrt \theta ,\,\,\,\,\,du = \cos \sqrt \theta \left( {\frac{1}{{2\sqrt \theta }}} \right)d\theta ,\,\,\,\,\,\,d\theta = \left( {\frac{{2\sqrt \theta }}{{\cos \sqrt \theta }}} \right)du \cr
& {\text{Write the integrand in terms of }}u \cr
& \int {\frac{{\cos \sqrt \theta }}{{\sqrt \theta {{\left( {\sin \sqrt \theta } \right)}^2}}}} d\theta = \int {\frac{{\cos \sqrt \theta }}{{\sqrt \theta {{\left( u \right)}^2}}}} \left( {\frac{{2\sqrt \theta }}{{\cos \sqrt \theta }}} \right)du \cr
& {\text{Simplifying}} \cr
& = \int {\frac{2}{{{u^2}}}du} \cr
& = - \frac{2}{u} + C \cr
& \cr
& {\text{Write in terms of }}\theta;{\text{ substitute }}\sin \sqrt \theta {\text{ for }}u \cr
& = - \frac{2}{{\sin \sqrt \theta }} + C \cr
& = - 2\csc \sqrt \theta + C \cr} $$
