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How do you integrate #int x^nsinx^(n-1)dx# using integration by parts?

1 Answer

Nov 7, 2015

I don't think you do. Here is the answer for #n=6# http://www.wolframalpha.com/input/?i=int+x%5E6sin%28x%5E5%29+dx

Explanation:

If you are not looking for an answer, you could, I suppose, use #u=x^2# and #dv=x^(n-2)sin(x^(n-1))dx# to get

#-1/(n-1)x^2cos(x^(n-1))-2/(n-1) int xcos(x^(n-1))dx#

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