How do you find the integral of #f(x)=x^nsinx^(n-1)# using integration by parts?

1 Answer
mason m
May 3, 2018

There's no closed form.

Explanation:

#I=intx^nsin(x^(n-1))dx#

Note that #intx^(n-2)sin(x^(n-1))dx# can be solved using the substitution #t=x^(n-1)#, since #dt=(n-1)x^(n-2)dx# and we see that the integral becomes

#intx^(n-2)sin(x^(n-1))dx =1/(n-1)int(n-1)x^(n-2)sin(x^(n-1))dx#

#=1/(n-1)intsin(t)dt=cos(x^(n-1))/(n-1)#

Motivated by this integrable function, let's rewrite #I#:

#I=intx^2x^(n-2)sin(x^(n-1))dx#

Now, let #dv=x^(n-2)sin(x^(n-1))dx# and let #u=x^2#. As we already determined, these mean that #v=cos(x^(n-1))/(n-1)# and it's easy to see that #du=2xdx#. Then:

#I=uv-intvdu=(x^2cos(x^(n-1)))/(n-1)-2/(n-1)intxcos(x^(n-1))dx#

Unfortunately, this integral has no closed form. Did you type your question right?

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