The following tables were taken from University of Pennsylvania's page about Calculus:

Trigonometric Substitution

Hyperbolic Substitution

As you can see, the forms $1+x^2$ and $x^2-1$ are repeated in the tables. How does one know when a trigonometric substitution is more suitable to a problem than a hyperbolic substitution and vice versa?

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    $\begingroup$ Note that despite the omission in the tables above, there is a trigonometric identity that can be used for the form $1 - x^2$, namely $\text{sech}^2 \theta = 1 - \tanh^2 \theta$. $\endgroup$ Commented Dec 20, 2014 at 16:16

3 Answers 3


This is a great question. I would add yet another option as well: Euler Substitution (which has various subtypes depending on parameters in the square root term).

In most cases, if one works, then all three will work. More generally, if you are performing an integral of the form $$ \int R(x, \sqrt{ax^2 + bx + c}) dx, $$ where $R(x,y)$ is a rational function of $x$ and $y$ (which means that it's a polynomial in $x$ and $y$ divided by another polynomial in $x$ and $y$), then all three techniques will generically work. Combining completing the square, partial fraction decomposition, and integration parts allows this to be stated more formally.

There is a problem in practice, though. Reducing a problem to the point where we need to perform partial fraction decompositions is great for numerics, but can be bad for exact solutions since we may need to decompose a polynomial whose factors we don't know exact forms for. (But we do know the partial fraction decomposition exists).

That's starting to get away from your question, though. A good starting rule is to always use trig instead of hyperbolic trig, since you are likely more familiar with rational integrals of trigonometric functions.

Some calculus instructors mention that if a trig sub yields integration of a power of $\sec x$ [which students often have trouble with because they often fall into the "tricky" double integration-by-parts category], you can instead do a hyperbolic trig sub and (often) get an "easier" integral. It is challenging to actually say it is "easier," since there is a strict algorithm for handling integrals of powers of $\sec x$ and for more general products of trig functions. So is it really easier to avoid it?

I once wrote a blog post on trig sub vs. hyperbolic trig sub vs. Euler sub. I perform one integral using all three techniques to see how the feel different, and talk a bit about motivation. If I were to boil everything down to a single heuristic, it's that there should be roughly a "Conservation of Difficulty" between these techniques.

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    $\begingroup$ Even though I'm a year and a bit late, this is a very helpful answer! +1 $\endgroup$
    – user141854
    Commented Aug 19, 2016 at 22:17

In my opinion, the following substitutions are enough:

appears in integrand substitution convention
1 $\sqrt{x^2+1}$ $x=\sinh\theta$ $\theta\in\mathbb R$
2 $\sqrt{x^2-1}$ $x=\pm\cosh\theta$ $\theta\geq0$
3 $\sqrt{1-x^2}$ $x=\sin\theta$ $-\frac{\pi}2\leq\theta\leq\frac{\pi}2$

$\sinh$ and $\cosh$ are better substitutions than $\tan$ and $\sec,$ respectively, as they are easier to differentiate and integrate, and have nicer principal domains. $\sin$ is a better substitution than $\tanh$ as it is easier to differentiate and integrate. The following examples illustrate this:

integrand substitution result
1 $\displaystyle\frac{x^2}{\sqrt{x^2+4}}$ $x=2\sinh\theta\quad\color{green}✔$ $4\int\sinh^2\theta\,\mathrm d\theta$
$\displaystyle\frac{x^2}{\sqrt{x^2+4}}$ $x=2\tan\theta\quad\color{red}✗$ $4\int\tan^2\theta\sec\theta\,\mathrm d\theta$
2 $\displaystyle\frac{\sqrt{x^2-4}}{x^2}\quad(x\leq-2)$ $x=-2\cosh\theta\quad\color{green}✔$ $-\int\tanh^2\theta\,\mathrm d\theta$
$\displaystyle\frac{\sqrt{x^2-4}}{x^2}\quad(x\leq-2)$ $x=2\sec\theta\quad\color{red}✗$ $-\int\tan^2\theta\cos\theta\,\mathrm d\theta$
3 $\sqrt{9-x^2}$ $x=3\sin\theta\quad\color{green}✔$ $9\int\cos^2\theta\,\mathrm d\theta$
$\sqrt{9-x^2}$ $x=3\tanh\theta\quad\color{red}✗$ $9\int\mathrm{sech^3}\,\theta\,\mathrm d\theta$
  • Trigonometric functions are basically Circular Functions which describe the horizontal and vertical positions of a point on a circle as a function of angle (cosine and sine).

  • Hyperbolic functions describe the same thing but can also be used to solve problem that can't be solved by Euclidean Geometry(where circular functions are sufficient).They can be used to describe Euclidean geometry but basically they are extension of it and used to solve problems from non-Euclidean geometry problems which arise in arise in many areas of Physics for example Relativity.

  • One useful example to make it clear to you would be derivation of Lorentz Transformation using Rotation of co-ordinates.Using circular functions will stop you at a point from solving the solution whereas using Hyperbolic functions leads to solving the equations leading to the famous Lorentz Transformations(One of the equations you use is from the table you have put up).

  • One another example from Mathematics is an integral.In these kinds of integral you may get answer using other substitutions but sometimes it is more natural to solve using Hyperbolic substitution for example the integral in this SE question: Integration Using Hyperbolic Substitution.

Read more @ http://mathworld.wolfram.com/HyperbolicFunctions.html to know how they extend the circular functions.

  • $\begingroup$ Do you realize the original question did not ask about the applications to physics? $\endgroup$ Commented Dec 20, 2014 at 16:18
  • $\begingroup$ This doesn't seem to answer OP's question, which asks how to determing a trig or hyperbolic trig substitution is more appropriate (presumably when evaluating integrals), not what trigonometric functions are or what applications they have. $\endgroup$ Commented Dec 20, 2014 at 16:18
  • $\begingroup$ @Travis ' How does one know when a trigonometric substitution is more suitable to a problem than a hyperbolic substitution and vice versa?' this is what he says.I just used a physics application to show him. $\endgroup$ Commented Dec 20, 2014 at 16:20
  • $\begingroup$ @DevarshRuparelia From the context of the link, I think OP is interested in evaluating integrals with trig/hyperbolic trig substitutions. $\endgroup$ Commented Dec 20, 2014 at 16:26
  • $\begingroup$ @Travis You are right but maybe he might just get the subtle difference of their use through this example. $\endgroup$ Commented Dec 20, 2014 at 16:28

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