trig/hyperbolic substitution integral

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 )^2

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.

Re: trig/hyperbolic substitution integral

Quote:

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.

Small mistake - it should be $\displaystyle \int \text{csch}^{4} y \ dy$ though I suspect that was a typo.

Write $\displaystyle \int \text{csch}^{4} y \ dy$ as $\displaystyle \int \text{csch}^{2}y \ \text{csch}^{2}y \ dy$ and now use integration by parts.

Recall that $\displaystyle \frac{d}{dx} \coth x = -\text{csch}^{2} x$.

Re: trig/hyperbolic substitution integral

Is there not a way to keep going only using trigonometric or hyperbolic substitutions. Or is it ok to use integration by parts when the initially question says 'Using a trigonometric or hyperbolic substitution, evaluate the following indenite integral'.

Kind regards

Re: trig/hyperbolic substitution integral