Answer
$$-\frac{1}{8} \cos 2 \theta-\frac{1}{16} \cos 4 \theta+\frac{1}{24} \cos 6 \theta+C $$
Work Step by Step
We make use of
$$\sin m x \cos n x=\frac{1}{2}[\sin (m-n) x+\sin (m+n) x]$$
Then
\begin{align*}
\int \sin \theta \sin 2 \theta \sin 3 \theta d \theta&=\int \frac{1}{2}(\cos (1-2) \theta-\cos (1+2) \theta) \sin 3 \theta d \theta\\
&=\frac{1}{2} \int(\cos (-\theta)-\cos 3 \theta) \sin 3 \theta d \theta\\
&=\frac{1}{2} \int \sin 3 \theta \cos \theta d \theta-\frac{1}{2} \int \sin 3 \theta \cos 3 \theta d \theta\\
&=\frac{1}{2} \int \frac{1}{2}(\sin (3-1) \theta+\sin (3+1) \theta) d \theta-\frac{1}{4} \int 2 \sin 3 \theta \cos 3 \theta d \theta\\
&=\frac{1}{4} \int(\sin 2 \theta+\sin 4 \theta) d \theta-\frac{1}{4} \int \sin 6 \theta d \theta\\
&=-\frac{1}{8} \cos 2 \theta-\frac{1}{16} \cos 4 \theta+\frac{1}{24} \cos 6 \theta+C
\end{align*}
