How do you evaluate the integral #int sqrt(1-1/x^2)#?

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Answer 1 · 2017-06-24T21:26:11

#I=intsqrt(1-1/x^2)dx#

Let's try to get the integrand to resemble #sqrt(1-sin^2theta)#. To do so, let #x=csctheta#. Then, #dx=-cscthetacotthetad theta#, and:

#I=intsqrt(1-1/csc^2theta)(-cscthetacotthetad theta)#

#color(white)I=intsqrt(1-sin^2theta)(-cscthetacottheta)d theta#

#color(white)I=-intcostheta1/sinthetacostheta/sinthetad theta#

#color(white)I=-intcot^2thetad theta#

Use #cot^2theta=csc^2theta-1#:

#I=int(1-csc^2theta)d theta#

#color(white)I=theta+cottheta#

Reverse the substitution #x=csctheta#:

#I=theta+sqrt(csc^2theta-1)#

#color(white)I=csc^-1x+sqrt(x^2-1)+C#