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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

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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.
User Jhickner
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