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