Prove

$\sin (a x+b) \cos (a x+b)=\frac{2 \sin (a x+b) \cos (a x+b)}{2}=\frac{\sin 2(a x+b)}{2}$

Let $2(a x+b)=t$

$\therefore 2 a d x=d t$

$\Rightarrow \int \frac{\sin 2(a x+b)}{2} d x=\frac{1}{2} \int \frac{\sin t d t}{2 a}$

$=\frac{1}{4 a}[-\cos t]+\mathrm{C}$

$=\frac{-1}{4 a} \cos 2(a x+b)+\mathrm{C}$