Question

How do you integrate `int x /sqrt(1 - x^2) dx` using trigonometric substitution?

Answer 1

`-(sqrt(1-x^2))+C`

Explanation:

Using

`sin^2x+cos^2x=1:.sin^2x=1-cos^2x`

`intx/(sqrt(1-x^2))dx`

substitue`" "x=sinu=>dx=cosudu`

we have:`" "intx/(sqrt(1-x^2))dx=int(sinu)/(sqrt(1-sin^2u))xx(cosu)du`

`" "=int(sinu)/(cancelcosu)xxcancel((cosu))du`

`=intsinudu=-cosu+C`

`=-(sqrt(1-x^2))+C`

this can also be integrated by inspection

`intx/(sqrt(1-x^2))dx=intx(1-x^2)^(-1/2)dx`

we note that a function of the derivative is outside the bracket ,so:

`d/(dx)((1-x^2)^(1/2))=1/2xx-2x(1-x)^(-1/2)=-x(1-x^2)^(-1/2)`

result follows

Answer 2

`int x/sqrt(1-x^2)dx = -sqrt(1-x^2)+C`

Explanation:

The integrand is defined only for `x in (-1,1)` so we can substitute `x=sint`, `dx=costdt` with `t in [-pi/2,pi/2]`

`int x/sqrt(1-x^2)dx = int (sintcost)/sqrt(1-sin^2t)dt=int (sintcost)/sqrt(cos^2t)dt = int (sintcost)/costdt= int sintdt=-cost+C`

In the same interval,

`cost= sqrt(1-x^2)`

So that:

`int x/sqrt(1-x^2)dx = -sqrt(1-x^2)+C`