Answer
$-2\displaystyle \sqrt{\frac{t}{a}}\cos\sqrt{at}+\frac{2}{a}\sin\sqrt{at}+C$
Work Step by Step
$\displaystyle \int\sin\sqrt{at}dt=$ substitute $\left[\begin{array}{ll}
u=\sqrt{at}, & u^{2}=at\\
& 2udu=adt\\
& \dfrac{2udu}{a}=dt
\end{array}\right]$
$\displaystyle \int\sin u\cdot\frac{2udu}{a} =\displaystyle \frac{2}{a}\int u\sin udu\quad $ ... by parts $= \displaystyle \frac{2}{a}\int vdw$
$\displaystyle \left[\begin{array}{ll}
v=u & dw=\sin udu\\
dv=du & w=-\cos u
\end{array}\right] =\frac{2}{a}[vw-\int wdv]$
$=\displaystyle \frac{2}{a}[-u\cos u-\int-\cos udu]$
$=\displaystyle \frac{2}{a}[-u\cos u+\sin u]+C$
... bring back t...
$=-\displaystyle \frac{2}{a}\sqrt{at}\cos\sqrt{at}+\frac{2}{a}\sin\sqrt{at}+C$
$=-2\displaystyle \sqrt{\frac{t}{a}}\cos\sqrt{at}+\frac{2}{a}\sin\sqrt{at}+C$
