Underlying question about u-substitution

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SUMMARY

The discussion centers on the application of u-substitution in the integral of the function f(x) = cos^4(x)sin(x) over the interval [0, π]. The initial substitution u = cos^4(x) led to an incorrect result of zero due to the non-injective nature of this substitution, as multiple x-values correspond to the same u-value. The correct approach, as indicated by the book, is to use u = cos(x), which yields a positive result. This highlights the importance of choosing appropriate substitutions in integral calculus.

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ehchandler
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So This problem arose from a homework problem, but this is a more underlying question about u-substitution since I understood completely how to do the problem the way the book wanted. The teacher was stumped too when I asked. There was an equation f(x)=cos^4(x)sin(x). We were supposed to take the integral of f(x) with [0,pi]. My inital intuition was just to make u=cos^4(x). Thus -du/4=cos^3(x)sin(x)dx. Then you get (-1/4)int(u^(1/4)) [1,1], which is obviously zero. This was wrong. I saw that this was strange, so I did way the book wanted, u=cos(x), but why are the two different. The bounds are in the positive range, so the ^4 shouldn't loose an answer or anything, and yet one comes out to be zero, while the other is a positive number. Why? Is there something underlying in the numbers themselves, or is it right in front of me. Thanks in advance.
 
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Your substitution u=cos^4(x) is not injective - multiple x-values are mapped to the same u-value. You would have to split the integral into several parts to do that.
 

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