Question

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

Answer 1

`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`