Question

What is `int_(0)^(6) (1+x)^3(12-2x)^(3/2)dx`?

Answer 1

`int_0^6(1+x)^3(12-2x)^(3/2)dx=(110304sqrt3)/55`

Explanation:

Let

`I=int_0^6(1+x)^3(12-2x)^(3/2)dx`

Apply the substitution `12-2x=u^2`:

`I=1/8int_0^sqrt12(14-u^2)^3u^4du`

Expand the cube:

`I=1/8int_0^sqrt12(2744u^4-588u^6+42u^8-u^10)du`

Integrate term by term:

`I=1/8[2744/5u^5-84u^7+14/3u^9-1/11u^11]_0^sqrt12`

Insert the limits of integration:

`I=(sqrt12)^5/1320(90552-13860(sqrt12)^2+770(sqrt12)^4-15(sqrt12)^6)`

Simplify:

`I=(12sqrt3)/55(90552-13860(12)+770(12)^2-15(12)^3)`

Hence

`I=(110304sqrt3)/55`