$int x^2 arcsin(x)dx =$ ?

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Answer 1 · 2016-12-28T15:25:40

$1/3x^3arcsin(x)+1/9x^2sqrt(1-x^2)+2/9sqrt(1-x^2)+C$

$d/(dx)(x^3 arcsin(x))=3x^2arcsin(x)+x^3/sqrt(1-x^2)$ then

$int x^2 arcsin(x)dx = 1/3x^3 arcsin(x)-1/3int x^3/sqrt(1-x^2)dx$

now $d/(dx)(x^2sqrt(1-x^2))=2 x sqrt[1 - x^2]-x^3/sqrt[1 - x^2]$

so

$int x^3/sqrt(1-x^2)dx=int 2 x sqrt[1 - x^2]dx-x^2sqrt(1-x^2)$

and

$int 2 x sqrt[1 - x^2]dx=-2/3 (1 - x^2)^(3/2)$

putting all together we have

$int x^2 arcsin(x)dx=1/3x^3arcsin(x)+1/9x^2sqrt(1-x^2)+2/9sqrt(1-x^2)+C$

int x^2 arcsin(x)dx = int x^2 arcsin(x)dx = ?