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OpenStudy (anonymous):
Find the integral of: 2x/3 sin(3x^2) dx
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OpenStudy (anonymous):
Take 2/3 out of the integral and integrate x/sin3x^2 which is = 1/6(log(sin(3x^2/2)))-log(cos93x^2/2)))+c
OpenStudy (anonymous):
Oops, I meant for there to be brackets around the 2x/3 - I get the idea though, thanks :)
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