To evaluate ∫ 3x2 cos (2x3 - 4) dx, it is necessary to let a. u = 6x. b. u = 6x2 c. u = 2x3 -4 d. u = 3x2

The correct answer is C, to let u = 2x^3 - 4.
This allows du = 6x^2 dx, and the integral will become:
integral of (1/2)(du) cos(u), which can be integrated in a straightforward manner.
The rule of thumb is to set the entire expression of the trigonometric function as a single variable if possible, to make it simpler.

Gotta be d the music of previous eras