please help, using trigonometric substitution

Question

∫9x²cos (4-3x³) dx

Answer 1

Let u = 4-3x³ Then, du = -9x²dx. So, 9x²dx = -du

So, ∫9x²cos(4-3x³)dx = -∫cosudu = -sinu + C = -sin(4-3x³) + C

Answer 2

∫9x²cos (4-3x³) dx

isn't this integral of cos(4-z)-dz,,,,z=3x³

which is -sin(4-z)+c

or ∫9x²cos (4-3x³) dx=-sin(4-3x³)+c

check d(-sin(4-3x³)+c)/x=-cos(4-x³)-*9x²=cos(4-x³)*9x²

this is simple fundamental thoerem of calculus not any trigonometirc substitution.

∫9x²cos (4-3x³) dx= -sin(4-3x³)+c