Answer
$$3 \sin \left(\frac{t}{6}\right)-\sin \left(\frac{t}{2}\right)+C$$
Work Step by Step
Use the formula
$$\int \sin a x \sin b x d x=\frac{\sin (a-b) x}{2(a-b)}-\frac{\sin (a+b) x}{2(a+b)}+C, \quad a^{2} \neq b^{2}$$
with $a=\dfrac{1}{3} ,\ \ \ b= \dfrac{1}{6}$, then
\begin{align*}
\int \sin \frac{t}{3} \sin \frac{t}{6} d t&=3 \sin \left(\frac{t}{6}\right)-\sin \left(\frac{t}{2}\right)+C
\end{align*}
