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July 29, 2014

July 29, 2014

Posted by **Hannah** on Friday, February 13, 2009 at 1:49am.

- Calculus -
**drwls**, Friday, February 13, 2009 at 1:56amLet 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)]

- Calculus -
**Reiny**, Friday, February 13, 2009 at 8:03amWouldn't it simply be

(1/6)sin(2x^3) + c ?

- Calculus -
**drwls**, Friday, February 13, 2009 at 8:39amThe 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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