# Calculus

to evaluate (integral) 3x^2 cos (2x^3-4) dx, it is necessary to let

A. u=3x^2
B. u=6x
C. u=2x^3-4
D. u=6x^2

1. 👍 0
2. 👎 0
3. 👁 102
1. C

then du = 2(3x^2) dx

1. 👍 0
2. 👎 0

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