Question

How do you integrate `int x^2sqrt(3-x)` using substitution?

Answer 1

`-2/35(3-x)^(3/2)(24+12x+5x^2)+C`.

Explanation:

To get rid of the Radical Sign, i.e., the Square Root, let us subst.

`3-x=t^2, or, x=3-t^2, so, dx=-2tdt`. Hence,

`I=intx^2sqrt(3-x)dx=int(3-t^2)^2*t*(-2t)dt`

`=-2int(9-6t^2+t^4)t^2dt`

`=-2int(9t^2-6t^4+t^6)dt`

`=-18intt^2dt+12intt^4dt-2intt^6dt`

`=-18*t^3/3+12*t^5/5-2*t^7/7`

`=-6t^3+12/5t^5-2/7t^7`

`=-2/35*t^3(105-42t^2+5t^4)`

`=-2/35*(t^2)^(3/2)(105-42t^2+5t^4)`

`=-2/35(3-x)^(3/2){105-42(3-x)+5(3-x)^2}`

`=-2/35(3-x)^(3/2)(24+12x+5x^2)+C`.

Enjoy Maths.!