If the solve the problem

We know $2 \sin A \cos B=\sin (A+B)+\sin (A-B)$

$\therefore \sin 4 x \cos 7 x=\frac{\sin 11 x+\sin (-3 x)}{2}$

We know $\sin (-\theta)=-\sin \theta$

$\therefore \sin (-3 x)=-\sin 3 x$

$\therefore$ The above equation becomes

$\Rightarrow \int \frac{1}{2}(\sin 11 x-\sin 3 x) d x$

$\Rightarrow \frac{1}{2}\left(\int \sin 11 x d x-\int \sin 3 x d x\right)$

We know $\int \sin \mathrm{ax} \mathrm{dx}=\frac{-1}{\mathrm{a}} \cos \mathrm{ax}+\mathrm{c}$

$\Rightarrow \frac{1}{2}\left(\frac{-1}{11} \cos 11 \mathrm{x}+\frac{1}{3} \cos 3 \mathrm{x}\right)$

$\Rightarrow \frac{11 \cos 3 \mathrm{x}-3 \cos 11 \mathrm{x}}{66}+c$