int x^2 arcsin(x)dx = ∫x2arcsin(x)dx= ?

1 Answer
Cesareo R.
Dec 28, 2016

1/3x^3arcsin(x)+1/9x^2sqrt(1-x^2)+2/9sqrt(1-x^2)+C13x3arcsin(x)+19x2√1−x2+29√1−x2+C

Explanation:

d/(dx)(x^3 arcsin(x))=3x^2arcsin(x)+x^3/sqrt(1-x^2)ddx(x3arcsin(x))=3x2arcsin(x)+x3√1−x2 then

int x^2 arcsin(x)dx = 1/3x^3 arcsin(x)-1/3int x^3/sqrt(1-x^2)dx∫x2arcsin(x)dx=13x3arcsin(x)−13∫x3√1−x2dx

now d/(dx)(x^2sqrt(1-x^2))=2 x sqrt[1 - x^2]-x^3/sqrt[1 - x^2]ddx(x2√1−x2)=2x√1−x2−x3√1−x2

so

int x^3/sqrt(1-x^2)dx=int 2 x sqrt[1 - x^2]dx-x^2sqrt(1-x^2)∫x3√1−x2dx=∫2x√1−x2dx−x2√1−x2

and

int 2 x sqrt[1 - x^2]dx=-2/3 (1 - x^2)^(3/2)∫2x√1−x2dx=−23(1−x2)32

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∫x2arcsin(x)dx=13x3arcsin(x)+19x2√1−x2+29√1−x2+C

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