How do you find the antiderivative of #int (x^3sinx) dx#?

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Answer 1 · 2016-10-16T08:30:49

#= - x^3 cos x + 3x^2 sin x + 6x cos x - 6 sin x + C#

#int x^3 sin x dx#

#= int x^3 (-cos x)^prime dx#

#= - x^3 cos x + int (x^3)^prime cos xdx#

#= - x^3 cos x + int 3x^2 cos xdx#

#= - x^3 cos x + int 3x^2 (sin x)^prime dx#

#= - x^3 cos x + ( 3x^2 sin x - int (3x^2)^prime sin x dx )#

#= - x^3 cos x + 3x^2 sin x - int 6x sin x dx #

#= - x^3 cos x + 3x^2 sin x - int 6x (- cos x)^prime dx #

#= - x^3 cos x + 3x^2 sin x - ( - 6x cos x + int (6x)^prime cos xdx )#

#= - x^3 cos x + 3x^2 sin x + 6x cos x - int 6 cos xdx #

#= - x^3 cos x + 3x^2 sin x + 6x cos x - 6 sin x + C#