BIến đổi f(x) về dạng:
$f(x)=\frac{3\sin x-\sin3x}{4} \sin3x=\frac{3}{4}\sin3x\sin x-\frac{1}{4}\sin^23x $
$=\frac{3}{8}(\cos2x-\cos4x)-\frac{1}{8}(1-\cos6x)=\frac{1}{8}(3\cos2x-3\cos4x+\cos6x-1)$
Khi đó:
$\int\limits f(x)dx=\frac{1}{8}\int\limits (3\cos2x-3\cos4x+\cos6x-1)dx$
$=\frac{1}{8}(\frac{3}{2}\sin2x-\frac{3}{4}\sin4x+\frac{1}{6}\sin6x-x)+C $