Question #9592b

1 Answer
Mar 3, 2017

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#