$\sin 3x\cos 4x$

: Let$I = \int {\sin 3x\cos 4x} dx$
$= \cfrac{1}{2}\int {\left[ {\sin \left( {3x + 4x} \right) + \sin \left( {3x - 4x} \right)} \right]} dx$

$= \cfrac{1}{2}\int {\left[ {\sin 7x + \sin \left( { - x} \right)} \right]} dx$

$= \cfrac{1}{2}\int {\left( {\sin 7x - \sin x} \right)dx} = \cfrac{1}{2}\left( {\cfrac{{ - \cos 7x}}{7}} \right) - \cfrac{1}{2}\left( { - \cos \,x} \right) + C$

$= - \cfrac{1}{{14}}\cos 7x + \cfrac{1}{2}\cos \,x + C$