Originally Posted by

**tylersmith7690** Using a trigonometric or hyperbolic substitution, evaluate the following indenite integral,

∫ 1 / ((x^2)-1)^(5/2) dx

my attempt:

let x= cosh y , dx/dy = sinh y , dx = sinhy dy

then (x^2-1) = (cosh^2 y) -1

= sinh^2 y

so ((x^2)-1)^(1/2) = sinh y

((x^2)-1)^(5/2) = ( sinh y ) ^5

so integral becomes sinh y / (sinh y)^5 dy

which is 1/ ( sinh y)^4 or ( cosech y )^4

I have tried to simplify these two integrals, one by putting cosh^2 y - sinh ^2 y in the numerator to replace the one in 1/ (sinh y)^4. But couldn't seem to get anywhere with the results. Is there some clever trick to apply to where I got down. I have spent way to long on this question and would like to see how someone else would work it out and give me some insight to there thinking.

Kind regards.