Answer
$$\,\int {\frac{{\cos \left( {2\sqrt x } \right)}}{{\sqrt x }}} dx = \sin \left( {2\sqrt x } \right) + C$$
Work Step by Step
$$\eqalign{
& {\text{Let }}\frac{d}{{dx}}\left[ {\sin \left( {2\sqrt x } \right)} \right] \cr
& {\text{using the chain rule }}\frac{d}{{dx}}\left[ {\sin u} \right] = \cos uu' \cr
& \,\,\,\,\,\,\,\,\,\,\,{\text{ }}\frac{d}{{dx}}\left[ {\sin \left( {2\sqrt x } \right)} \right] = \cos \left( {2\sqrt x } \right)\left( {2\sqrt x } \right)' \cr
& {\text{Differentiating}} \cr
& \,\,\,\,\,\,\,\,\,\,\,{\text{ }}\frac{d}{{dx}}\left[ {\sin \left( {2\sqrt x } \right)} \right] = 2\cos \left( {2\sqrt x } \right)\left( {\frac{1}{{2\sqrt x }}} \right) \cr
& {\text{Simplifying}} \cr
& \,\,\,\,\,\,\,\,\,\,\,{\text{ }}\frac{d}{{dx}}\left[ {\sin \left( {2\sqrt x } \right)} \right] = \frac{{\cos \left( {2\sqrt x } \right)}}{{\sqrt x }} \cr
& {\text{Thus}}{\text{,}}\,\,\,\,{\text{a corresponding integration formula is}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\int {\frac{{\cos \left( {2\sqrt x } \right)}}{{\sqrt x }}} dx = \sin \left( {2\sqrt x } \right) + C \cr} $$
