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

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Answer 1 · 2018-08-06T23:39:58

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

#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#

\0/ Here's our answer !