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

Question

4707 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-05T15:19:32

answer
#int x^nsinx^(n)dx=[1/(n+1)*x^(n+1)]*sin(x^n)-int[n*x^(2n)*cosx^n]/(n+1)*dx#

#int x^nsinx^(n)dx#

suppose:

#u=sin(x^n)#

#du=cos(x^n)*n*x^(n-1)*dx#

#dv=x^n*dx#

#v=1/(n+1)*x^(n+1)#

#intu*dv=uv-intv*du#

#int x^nsinx^(n)dx=[1/(n+1)*x^(n+1)]*sin(x^n)-int[1/(n+1)*x^(n+1)][cos(x^n)*n*x^(n-1)]*dx#

#int x^nsinx^(n)dx=[1/(n+1)*x^(n+1)]*sin(x^n)-int[n*x^(2n)*cosx^n]/(n+1)*dx#