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

2 Answers
Dec 21, 2016

#-(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

Dec 21, 2016

#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#