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

Question

12773 views around the world

You can reuse this answer
Creative Commons License

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#