Answer
$$\frac{1}{2} \cos \theta-\frac{1}{22} \cos 11 \theta+C $$
Work Step by Step
Since $$\sin m \cos n =\frac{1}{2}[ \sin(m-n)+ \sin(m+n)]$$
Then
\begin{align*}
\int \sin 5 \theta \cos 6 \theta d \theta&=\frac{1}{2} \int(\sin (-\theta)+\sin (11 \theta)) d \theta\\
&=\frac{1}{2} \int \sin (-\theta) d \theta+\frac{1}{2} \int \sin (11 \theta) d \theta\\&=\frac{1}{2} \cos (-\theta)-\frac{1}{22} \cos 11 \theta+C\\
&=\frac{1}{2} \cos \theta-\frac{1}{22} \cos 11 \theta+C
\end{align*}
