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Im doing substitution right now in calcalus and im kinda confused on how to do this problem. {integral}sqrt(x)sin^2(x^(3/2)-1)dx. u=x^(3/2)-1 Im doing substitution right now in calcalus and im kinda confused on how to do this problem. {integral}sqrt(x)sin^2(x^(3/2)-1)dx. u=x^(3/2)-1 @Mathematics
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\[\int\limits_{}^{}\sqrt{x}\sin^2(x^{3/2}-1)dx\] \[u=x^{3/2}-1\] \[du=(3/2)\sqrt{x}\] \[(3/2)\int\limits_{}^{}\sin^2(u)du\] Evaluate this integral and substitute \[u = x^{3/2}\]
so how did you get rid of the sqrt(x)?
It "disappears" because it is part of du.
ahhh ok i see it tyvm
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