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

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Answer 1 · 2018-05-05T15:19:32

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

$int x^nsinx^(n)dx$

suppose:

$u=sin(x^n)$

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

$dv=x^n*dx$

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

$intu*dv=uv-intv*du$

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

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

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