Answer
$$\int \sin (-7 x) \cos 6 x d x = \frac{1}{26}\cos (13 x)+\frac{1}{2} \cos (x) +c$$
Work Step by Step
$$
\int \sin (-7 x) \cos 6 x d x
$$
Since $$
\sin m x \cos n x=\frac{1}{2}(\sin [(m-n) x]+\sin [(m+n) x])
$$
Then
\begin{align*}
\int \sin (-7 x) \cos 6 x d x&=\frac{1}{2}\int\left( \sin (-13 x)+ \sin (-x)\right) d x\\
&=\frac{1}{2}\int\left(-\sin (13 x)- \sin (x)\right) d x\\
&=\frac{1}{2}\left(\frac{1}{13}\cos (13 x)+ \cos (x)\right)+c\\
&= \frac{1}{26}\cos (13 x)+\frac{1}{2} \cos (x) +c
\end{align*}
