Recall that
[tex]\cos(x\pm y)=\cos x\cos y\mp\sin x\sin y[/tex]
This means
[tex]\cos12x=\cos4x\cos8x-\sin4x\sin8x[/tex]
[tex]\cos4x=\cos4x\cos8x+\sin4x\sin8x[/tex]
which then means
[tex]\sin4x\sin8x=\dfrac{\cos4x-\cos12x}2[/tex]
So,
[tex]\displaystyle\int\sin4x\sin8x\,\mathrm dx=\frac12\int(\cos4x-\cos12x)\,\mathrm dx[/tex]
[tex]=\dfrac12\left(\dfrac14\sin4x-\dfrac1{12}\sin12x\right)+C[/tex]
[tex]=\dfrac18\sin4x-\dfrac1{24}\sin12x+C[/tex]
