How do you integrate $int x^3sinx$ by parts?

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Answer 1 · 2018-01-06T05:57:04

$I=-x^3cosx+3x^2sinx+6xcosx-6sinx+"C"$

The formula for Integration by Parts (IBP): $int$ $u$ $dv=uv-int$ $v$ $du$

Let $color(red)(u_1=x^3;dv_1=sinx$

Thus, $color(red)(du_1=3x^2dx;v_1=-cosx$

$I=color(red)([x^3][-cosx])-int[-cosx][3x^2dx]$

$I=color(red)(-x^3cosx)+int3x^2cosxdx$

Apply IBP again:

Let $color(blue)(u_2=3x^2;dv_2=cosx$

Thus, $color(blue)(du_2=6xdx;v_2=sinx$

$I=color(red)(-x^3cosx)+{color(blue)([3x^2][sinx])-int[sinx][6xdx]}$

$I=color(red)(-x^3cosx)color(blue)(+3x^2sinx)-int6xsinxdx$

Apply IBP once more:

Let $color(green)(u_3=6x;dv_3=sinx$

Thus, $color(green)(du_3=6dx;v_3=-cosx$

$I=color(red)(-x^3cosx)color(blue)(+3x^2sinx)-{color(green)([6x][-cosx])-int[-cosx][6dx]}$

$I=color(red)(-x^3cosx)color(blue)(+3x^2sinx)-{color(green)(-6xcosx)+6intcosxdx}$

Since $intcosxdx=color(purple)(sinx+"C"$

$I=color(red)(-x^3cosx)color(blue)(+3x^2sinx)-{color(green)(-6xcosx)+color(purple)(6[sinx]}}+"C"$

$I=color(red)(-x^3cosx)color(blue)(+3x^2sinx)color(green)(+6xcosx)color(purple)(-6sinx+"C")$

How do you integrate int x^3sinxint x^3sinx by parts?