Question

1/(e^x-e^-x) integral

Answer 1

dxex+e−x∫dxex+e−x

Multiplying numerator and denominator by exex

∫exe2x+1dx∫exe2x+1dx

Put ex=tex=t therefore dt=exdxdt=exdx

∫dt1+t2∫dt1+t2

=arctant+C=arctan⁡t+C

Putting in original value, final answer is

=arctan(ex)+C=arctan⁡(ex)+C

Well, another way to do this is using some Complex analysis.

Using Euler’s Identity

eix=cosx+isinxeix=cos⁡x+isin⁡x

ex=cosxi+isinxiex=cos⁡xi+isin⁡xi

ex=cos(ix)−isin(ix)ex=cos⁡(ix)−isin⁡(ix)

SimilarlySimilarly

e−x=cos(ix)+isin(ix)e−x=cos⁡(ix)+isin⁡(ix)

Our integral then transform into

∫dxcosix−isinix+cosix+isinix∫dxcos⁡ix−isin⁡ix+cos⁡ix+isin⁡ix

=12∫sec(ix)dx=12∫sec⁡(ix)dx

Using Substitution of ix=t⟹dx=(1/i)dtix=t⟹dx=(1/i)dt

We get the result

12iln(sec(ix)+tan(ix))+C12iln⁡(sec⁡(ix)+tan⁡(ix))+C

Which can be written in Hyperbolic Trigonometric Functions as

12iln(sechx+tan(ix))+C12iln⁡(sechx+tan⁡(ix))+C

Note: The function in the question itself can be written as Hyperbolic Trigonometric Function

sechx2sechx2