Question

Question #9592b

Answer 1

To find the indefinite integral, let `x = sect`,

so that `x^2 -1 = tan^2x` and `dx = sect tant dt`.

Upon substitution we will get

`int dx/(x^2-1)^(3/2) = int (sec t tant)/tan^3t dt`

`= int sect cot^2 t dt`

`= int csct cot t dt = -csc t +C`

Since `x = sect`, we have `cos t = 1/x`, so

`sin t = sqrt(x^2-1)/x` and

`-csct +C= -1/sint =-x/sqrt(x^2-1)+C`

If the integrand should be `abs(x^2-1)`, Then for `a < 1`, use

`abs(x^2-1) = (1-x^2)`.

In this case, substitute `x = sint` and integrate to get

`x/sqrt(x^2-1)+C`