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

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Answer 1 · 2016-09-14T09:04:45

#= -ln((sqrt(1-x^2)+ 1)/x )+ sqrt(1-x^2) + C#

we can try the obvious #x = sin y, dx/dy = cos y# giving

#int sqrt(1-x^2)/x dx#

#= int sqrt(1-sin^2 y )/(sin y) cos y \ dy#

#= int cos^2 y/(sin y) \ dy#

#= int (1-sin^2 y)/(sin y) \ dy#

#= int csc y - sin y \ dy#

using standard integrals
#= -ln(cot y+csc y )+ cos y + C#

Now converting back to x meaning:
#sin y = x, csc y = 1/x, cot y = (sqrt(1-x^2))/x#

#= -ln((sqrt(1-x^2)+ 1)/x )+ sqrt(1-x^2) + C#