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 $x=csctheta$ :

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

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

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