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

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Answer 1 · 2016-10-16T08:30:49

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

$int x^3 sin x dx$

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

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

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

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

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

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

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

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

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

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

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