Question

How do you integrate `int (x^2 -x) / sqrt(9 - x^2) dx` using trigonometric substitution?

Answer 1

`I=9/2sin^-1(x/3)-x/2sqrt(9-x^2)+sqrt(9-x^2)+c`

Explanation:

We know that,

`color(red)((1)sin^2x=(1-cos2x)/2)`

`color(blue)((2)intcosxdx=sinx+c)`

`color(violet)((3)intsinxdx=-cosx+c)`

Here,

`I=int(x^2-x)/sqrt(9-x^2)dx`

Subst. `x=3sinu=>dx=3cosudu`

`=>x^2=9sin^2u and sinu=x/3=>u=sin^-1(x/3)`

So,

`I=int(9sin^2u-3sinu)/sqrt(9-9sin^2u)xx3cosudu`

`=int(9sin^2u-3sinu)/(cancel(3cosu))xxcancel(3cosu)du`

`=9intsin^2udu-3intsinudu...tocolor(red)(Apply(1)`

`=9int(1-cos(2u))/2du-3intsinudu...toApplycolor(blue)((2)) andcolor(violet)((3))`

`=9/2[u-sin(2u)/2]-3(-cosu)+c`

`=9/4[2u-sin2u]+3cosu+c`

Subst. back , `sinu=x/3 and u=sin^-1(x/3)`

`I=9/4[2sin^-1(x/3)-2sinucosu]+3sqrt(1-sin^2u)+c`

`=9/4xx2[sin^-1 (x/3)-sinusqrt(1-sin^2u)]+3sqrt(1-x^2/9)+c`

`=9/2[sin^-1(x/3)-(x/3)sqrt(1-x^2/9)]+3sqrt(9-x^2)/3+c`

`=9/2sin^-1(x/3)-9/2xxx/3sqrt(9-x^2)/3+sqrt(9-x^2)+c`

`I=9/2sin^-1(x/3)-x/2sqrt(9-x^2)+sqrt(9-x^2)+c`