How do you integrate #int x sin sqrtx dx # using integration by parts?

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Answer 1 · 2016-01-24T14:19:10

let's #u = sqrt(x)#

#du = 1/(2sqrt(x))dx#

#2udu = dx#

#u^2=x#

so you have

#2intu^3sin(u)du#

And integrate by part three time and you have

#2[(3 (u^2-2) sin(u)-u (u^2-6) cos(u))]+C#