Main Question

I have some beef I'd like to let loose for the following integral:

$$\int \frac{1}{e^x-e^{-x}} dx$$

This is equivalent to the following, with the appropriate substitution of $t = e^x$:

$$\int \frac{1}{t^2-1} dt$$

I'm going to omit the proof for this but it's quite trivial and not part of my question. Now, the route my lecturer used was integration by partial fractions, and that's well and good, but my beef with the question has to do my following flawed processes, and why they aren't valid:

  • Recall $cosh^2\ x - sinh^2\ x = 1$
  • Hence $-sinh^2\ x - 1 = -cosh^2\ x$
  • Let $-sinh\ x = t$ so $sinh^2\ x = t^2$
  • $\frac{dt}{dx} = -cosh\ x$
  • $dx = \frac{-1}{cosh\ x}$

Thus the integral is now:

$$\int \frac{1}{sinh^2x\ cosh\ x} dx$$

This is obviously utterly wrong. The mistake must've arisen from changing the differential variable dx to dt. However, what if I did this?

  • $t = e^x$.
  • Let $e^x = cosh\ x$

What would I have to do now to allow myself to plug in $cosh^2\ x$ for $e^{2x}$? Why, if I cannot do this, is this not allowed?


closed as unclear what you're asking by Did, Claude Leibovici, user91500, Daniel W. Farlow, TomGrubb Jun 14 '17 at 22:07

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  • $\begingroup$ So we agree that $\int 1/(t^2 + 1) dt = \arctan(t) + C$ right ? $\endgroup$ – Zubzub Jun 12 '17 at 7:54
  • $\begingroup$ Regarding your first question, I think none of the substitutions you did are correct. For the extra question, please, write it in a new question; it will help you getting a better answer. $\endgroup$ – AugSB Jun 12 '17 at 8:09
  • $\begingroup$ @Zubzub I recall that being correct, but why is my statement wrong for why I can't use a cosh identity there? $\endgroup$ – sangstar Jun 12 '17 at 8:16
  • $\begingroup$ "This is equivalent to the following, with the appropriate substitution of t=ex:" Absolutely not. $\endgroup$ – Did Jun 12 '17 at 8:19
  • 1
    $\begingroup$ The integral is $\int e^{-x}\frac{1}{1-1/e}dx$ $\endgroup$ – Aaron Jun 12 '17 at 8:19

Just as a small remark, not really answering the question: It's much easyer to work out the integral by observing $$\frac{1}{e^x - e^{x-1}} = e^{-x} \frac{1}{1 - e^{-1}}.$$ Since the latter factor is constant, it's sufficient to know the primitive of $e^{-x}$ which should be well-known.

  • $\begingroup$ That's a very good point, wish I spotted that. $\endgroup$ – sangstar Jun 12 '17 at 15:46

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