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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$\displaystyle sin(x^3) x^2 dx $* With the substitution you proposed you have $\displaystyle u = x^3 $ $\displaystyle \frac{du}{3} = x^2 dx $ And the expression becomes $\displaystyle sin(u) \frac{du}{3} $
I'm sorry but why is it dU/3? where did you get that?
$\displaystyle d(u) = du $ $\displaystyle d(u) = d(x^3) = 3x^2 dx = du $ If you divide the two last term by 3 you get $\displaystyle x^2 dx = \frac{du}{3} $
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