If we were to take [tex]y=2x^3-4[/tex], then we have [tex]\mathrm dy=6x^2\,\mathrm dx[/tex]. So in the integral, we could write
[tex]\displaystyle\int3x^2\cos(2x^3-4)\,\mathrm dx=\int2(3x^2)\cos(2x^3-4)\,\mathrm dx=2\int\cos y\,\mathrm dy=2\sin y+C[/tex]
Then back-substituting to get the antiderivative in terms of [tex]x[/tex], we have
[tex]\displaystyle\int3x^2\cos(2x^3-4)\,\mathrm dx=2\sin(2x^3-4)+C[/tex]
