What is intx/sqrt(1-x^2)x1x2 ?

1 Answer
Aug 6, 2018

intx/sqrt(1-x^2)dx=-sqrt(1-x^2)+Cx1x2dx=1x2+C, C in RR

Explanation:

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

let y=sqrt(1-x^2)

y^2=1-x^2
y^2-1=-x^2
x=sqrt(1-y^2)
dx=1/2*-2y*1/sqrt(1-y^2)dy

dx=-y/sqrt(1-y^2)dy

So:

I=int((sqrt(1-y^2))*(-y)/(sqrt(1-y^2)))/ydy

=-intcancel(y/y)*cancel(sqrt(1-y^2)/sqrt(1-y^2))dy

=-int1dy

=-y

=-sqrt(1-x^2)+C, C in RR

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