the problem is Sin(x^3) x^2 dx
u = x^3
du = 3x^2
I noticed when my teacher was solving this out her final answer was -1/3 cos(x^3) + C
My question is where did the x^2 go and where did she pull out the 1/3 from?
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With the substitution you proposed you have And the expression becomes
I'm sorry but why is it dU/3? where did you get that?
If you divide the two last term by 3 you get
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