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How do you integrate #int x /sqrt(1 - x^2) dx# using trigonometric substitution?
2 Answers
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#
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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
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,
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#cost= sqrt(1-x^2)#
So that:
#int x/sqrt(1-x^2)dx = -sqrt(1-x^2)+C#
Related topic
Integration by Trigonometric Substitution
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