Answer
\[2\sqrt x \sin \sqrt x + 2\cos \sqrt x + C\]
Work Step by Step
\[\begin{gathered}
\int {\cos \sqrt x } dx \hfill \\
{\text{Let }}z = \sqrt x ,{\text{ }}dz = \frac{1}{{2\sqrt x }}dx = \frac{1}{{2z}}dx{\text{ }} \Rightarrow dx = 2zdz,{\text{ then}} \hfill \\
\int {\cos \sqrt x } dx = \int {2z\cos z} dz \hfill \\
{\text{Integrate by parts, }} \hfill \\
{\text{let }}u = 2z{\text{ }}du = 2dz \hfill \\
dv = \cos zdz{\text{ }}v = \sin z \hfill \\
{\text{Therefore,}} \hfill \\
\int {2z\cos z} dz = 2z\sin z - \int {\sin z\left( 2 \right)} dz \hfill \\
{\text{ }} = 2z\sin z - 2\int {\sin z} dz \hfill \\
{\text{ }} = 2z\sin z + 2\cos z + C \hfill \\
{\text{Back - substitute }}z = \sqrt x \hfill \\
{\text{ }} = 2\sqrt x \sin \sqrt x + 2\cos \sqrt x + C \hfill \\
\end{gathered} \]
