Question

What is the integrate x^2*square root (4-x^2)?

Answer 1

`int x^2sqrt(4-x^2)*dx`

=`2arcsin(x/2)-1/4*(2x-x^3)*sqrt(4-x^2)+C`

Explanation:

`int x^2sqrt(4-x^2)*dx`

After using `x=2sinu` and `dx=2cosu*du` transforms, this integral became,

`int (2sinu)^2*2cosu*2cosu*du`

=`int 16(sinu*cosu)^2*du`

=`int 4(sin2u)^2*du`

=`int (2-2cos4u)*du`

=`2u-1/2sin4u+C`

=`2u-1/2*4sinu*cosu*cos2u+C`

=`2u-2sinu*cosu*cos2u+C`

=`2u-2sinu*cosu*(1-2(sinu)^2)+C`

After using `x=2sinu`, `sinu=x/2`, `cosu=sqrt(4-x^2)/x` and `u=arcsin(x/2)` inverse transforms, I found

`int x^2sqrt(4-x^2)*dx`

=`2arcsin(x/2)-2*x/2*sqrt(4-x^2)/2*(1-2*x^2/4)+C`

=`2arcsin(x/2)-1/4*(2x-x^3)*sqrt(4-x^2)+C`