Answer
$$2\sqrt x \sin \sqrt x + 2\cos \sqrt x + C$$
Work Step by Step
$$\eqalign{
& \int {\cos \sqrt x } dx \cr
& {\text{Let }}t = \sqrt x ,{\text{ }}{t^2} = x,{\text{ }}2tdt = dx \cr
& {\text{Use the substitution}} \cr
& \int {\cos \sqrt x } dx = \int {\cos t} \left( {2t} \right)dt \cr
& = 2\int {t\cos t} dt \cr
& {\text{Integrate by parts,}} \cr
& {\text{Let }}u = t,{\text{ }}du = dt \cr
& dv = \cos tdt,{\text{ }}v = \sin t \cr
& 2\int {t\cos t} dt = 2\left( {t\sin t - \int {\sin tdt} } \right) \cr
& 2\int {t\cos t} dt = 2\left( {t\sin t - \left( { - \cos t} \right)} \right) + C \cr
& 2\int {t\cos t} dt = 2t\sin t + 2\cos t + C \cr
& {\text{Back - substitute }}t = \sqrt x \cr
& \int {\cos \sqrt x } dx = 2\sqrt x \sin \sqrt x + 2\cos \sqrt x + C \cr} $$
