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

Question

5491 views around the world

You can reuse this answer
Creative Commons License

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")#