2024/017) integrate sin ax cos bx

We have $\sin \, ax + \cos \, bx = \frac{1}{2} (\sin (a+b) x + \sin (a-b) x$

Knowing that 

$\frac{d}{dx} \ cos \, nx = - n \sin \, nx$

Hence $\int  (\sin \, ax  \cos \, bx) dx  = \int (\frac{1}{2} (\sin (a+b) x + \sin (a-b) x)dx$ 

= $\frac{\cos(a+b)x}{a+b} - \frac{\cos(a-b)x}{a-b} + C$