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When I was a Calculus II student many (about 4800) moons ago, our professor taught us an alternative to trig sub. For example, if we have $$ \int \frac{dx}{x^2\sqrt{x^2 - 9}}, $$ we would evaluate with trig sub by letting $x = 3\sec\theta$ and we'd get: \begin{align} \int \frac{dx}{x^2\sqrt{x^2 - 9}} &= \int\frac{3\sec\theta\tan\theta \, d\theta}{9\sec^2\theta\sqrt{9\sec^2\theta - 9}}\\[0.3cm] &= \frac{1}{3}\int\frac{\tan\theta\,d\theta}{\sec\theta \cdot 3\tan\theta}\\[0.3cm] &= \frac{1}{9} \int \cos\theta \, d\theta\\[0.3cm] &= \frac{1}{9} \sin\theta + C\\[0.3cm] &= \frac{\sqrt{x^2-9}}{9x} + C \end{align}

The alternative method he showed us goes like this for this problem: \begin{align} \int\frac{dx}{x^2\sqrt{x^2 - 9}} &= \int \frac{dx}{x^2\sqrt{x^2(1 - 9x^{-2})}}\\[0.3cm] &= \int\frac{dx}{x^3\sqrt{1 - 9x^{-2}}}\\[0.3cm] &= \int\frac{x^{-3} \, dx}{\sqrt{1 - 9x^{-2}}} \end{align} Yes, I know that technically $\sqrt{x^2} = |x|$. But trig sub also comes with domain restrictions.

Now let $u = 1 - 9x^{-2}$. Then $du = 18x^{-3} \, dx$ and we have: \begin{align} \int\frac{x^{-3} \, dx}{\sqrt{1 - 9x^{-2}}} &= \frac{1}{18}\int\frac{du}{\sqrt{u}}\\[0.3cm] &= \frac{1}{18} \cdot 2\sqrt{u} + C\\[0.3cm] &= \frac{\sqrt{1 - 9x^{-2}}}{9} + C \end{align}

Same answer, different form. My question has two parts, sort of (since an answer to #1 could point to an answer to #2).

  1. Does anyone know the history of this method?
  2. Is this method perfectly interchangeable with trig sub? In other words, can an integral be done using this method iff it can also be done using trig sub?

IIRC, which I may not because of all the moons, our professor said there's no "general formula" to get this to work. You just kind of have to eyeball it and try it. But I don't remember him saying whether or not it would work all the time.

Thanks!

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  • $\begingroup$ Yes they are perfectly interchangable because the domain for the substitued = range for the new substitution. $\endgroup$ – user2277550 Mar 24 '16 at 17:53
  • $\begingroup$ Suppose we were to integrate from $-5$ to $-3$. Will the end answers be the same?.. $\endgroup$ – Qwerty Mar 24 '16 at 17:59
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    $\begingroup$ You are 400 years old... You should have asked the questione to Cauchy or, more recently, to Lebesgue. $\endgroup$ – N74 Mar 24 '16 at 21:47
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set $$\sqrt{x^2-9}=t+x$$ then you will get $$x=\frac{t^2+9}{2t}$$ then we will have $$dx=\frac{1}{2}\frac{t^2-9}{t^2}dt$$

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  • $\begingroup$ Sorry, I don't understand how this addresses the question. $\endgroup$ – user307169 Mar 24 '16 at 18:29
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    $\begingroup$ @tilper: It points to a family of non-trig substitutions (the Euler substitutions) that can be used to solve such problems. $\endgroup$ – André Nicolas Mar 24 '16 at 18:32
  • $\begingroup$ Ok, thanks. I like how one of the Wiki references on the Euler substitution page is the book I'm planning to read next. $\endgroup$ – user307169 Mar 24 '16 at 19:06
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While I do not know the history on this type of substitution, an early reference were I have seen such a thing used can be found in "A treatise on the integral calculus. Containing the integration of explicit functions of one variable; together with the theory of definite integrals and of elliptic functions" by J. Hymer which was published in 1835. Starting on page 8 a number of examples using such a substitution are given but unfortunately the author gives no reference to the use of such a technique.

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  • $\begingroup$ Thanks, I'll try to look that book up. $\endgroup$ – user307169 Mar 25 '16 at 5:25
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    $\begingroup$ A copy of the complete book can be found at: www.archive.org $\endgroup$ – omegadot Mar 25 '16 at 5:27

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