Posted by Hannah on Friday, February 13, 2009 at 1:49am.
Let w = 2x^3
6x^2 dx = dw
The integral becomes the integral of
(1/6)w cos w, which can be solved using integration by parts to give
(1/6)[cos w + w sin w]
= (1/6)[cos(2x^3) + 2x^2 sin(2x^3)]
Wouldn't it simply be
(1/6)sin(2x^3) + c ?
The derivative of Reiny's answer is
(1/6)cos(2x^3)*6x^2 = x^2 cos(2x^3), so I must have made a mistake somewhere.
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