Answer
$2(\sqrt{t}\sin\sqrt{t}+\cos\sqrt{t})+C$
Work Step by Step
$\displaystyle \int\cos\sqrt{t}dt=\quad\left[\begin{array}{ll}
r=\sqrt{t}, & r^{2}=t\\
& 2rdr=dt
\end{array}\right]$
$=\displaystyle \int 2r\cos rdr$
by parts, $\displaystyle \left[\begin{array}{lll}
u=2r & dv=\cos rdr & \\
du=2dr & v=\sin r &
\end{array}\right],\quad\int udv=uv-\int vdu$
$=2r\displaystyle \sin r-2\int\sin rdr$
$=2r\sin r+2\cos r+C$
... bring back t
$=2(\sqrt{t}\sin\sqrt{t}+\cos\sqrt{t})+C$
