0
$\begingroup$

I'm asked to integrate $\int{dx\over x^2 - 9}$ using hyperbolic substitution. Using the relation cosh^2-1 = sinh^2, I let x = 3cosh(u), and through simplification, arrived at $\int$ ${1\over 3 csch(u)}du$. I integrated and got $1\over 3$ln(tanh(${u \over 2}$)+c. This is where I'm stuck, as I don't know how to convert back into terms of 'x' from 'u'. If anyone could help me out, it would be much appreciated!

Thank you!

$\endgroup$
3
  • $\begingroup$ Look here to learn $\endgroup$
    – suomynonA
    Oct 20, 2016 at 4:21
  • $\begingroup$ you almost have it, but you have to enclose mathjax in dollar signs, for example $1+1\ne2$ shows up as $1+1\ne2$ $\endgroup$
    – suomynonA
    Oct 20, 2016 at 4:38
  • $\begingroup$ I think I've got it now, I was testing it to see if clicking "save" applied the edits (I quickly learned it did not haha) $\endgroup$
    – Ben
    Oct 20, 2016 at 4:42

2 Answers 2

0
$\begingroup$

best method is to use partial fractions $$\begin{align} \int\frac{dx}{x^2-9}dx\\=\frac{1}{6}\int\left(\frac{1}{x-3}-\frac{1}{x+3}\right)dx\\ =\frac{1}{6}(\ln|x-3|-\ln|x+3|)\\ =\frac{1}{6}\ln|(x-3)/(x+3)|+C \end{align}$$

$\endgroup$
1
  • $\begingroup$ I realize that, this question requires me to use hyperbolic substitution for practice. Thank you though! $\endgroup$
    – Ben
    Oct 20, 2016 at 5:10
0
$\begingroup$

The most straightforward way to go is to apply inverse hyperbolic functions (similar to arcsine and friends). Then \begin{align} x &= 3 \cosh u \\ \frac{x}{3} &= \cosh u \\ \cosh^{-1} \left( \frac{x}{3} \right) &= u \text{.} \end{align} Analogous to the inverse (circular) trigonometric functions, one should remember that this inverse function has a restricted domain, $[1,\infty)$, which can cause similar surprises.

Then it's useful to know the alternative forms of the arc-hyperbolic functions. For arc hyperbolic cosine, you want $$ \cosh^{-1} \alpha = \ln(\alpha + \sqrt{\alpha^2 - 1}) $$

$\endgroup$
6
  • $\begingroup$ That makes sense, thank you so much! $\endgroup$
    – Ben
    Oct 20, 2016 at 5:10
  • $\begingroup$ I've followed your steps to the tee and understand all of them, but I'm not sure how to tie it in to tan(${u\over 2] $\endgroup$
    – Ben
    Oct 20, 2016 at 6:14
  • $\begingroup$ ** tanh(${u \over 2}$ sorry. Could you explain that to me? $\endgroup$
    – Ben
    Oct 20, 2016 at 6:17
  • $\begingroup$ Actually, I'm thinking I could use a hyperbolic half angle ID to pull it out of tanh and into cosh. Then, I could use the cosh value to relate it back into x. Does that seem logical? $\endgroup$
    – Ben
    Oct 20, 2016 at 6:21
  • $\begingroup$ You seem to be asking for $\tanh \frac{u}{2} = \tanh \frac{\cosh^{-1} (x/3)}{2}$. Perhaps you mean to ask about half angle formulas for hyperbolic functions, for instance $\tanh \frac{u}{2} = \frac{\sinh u}{\cosh(u)+1} = \frac{\sinh(\cosh^{-1} (x/3))}{\cosh(\cosh^{-1} (x/3))+1}$. Now set up your triangles the same way you would in trigonometry to deal with the analogous expressions $\sin \cos^{-1}(x/3)$ and $\cos \cos^{-1}(x/3)$. (Note that these triangles have a point on a hyperbola, not on the unit circle.) $\endgroup$ Oct 21, 2016 at 8:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .