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In an old Italian calculus problem book, there is an example presented:


The solution given uses the strange substitution $$x=\frac{1}{1-u}$$

Some preliminary work in trying to determine the motivation as to why one would come up with such an odd substitution yielded a right triangle with hypotenuse $x$ and leg $x-1;$ determining the other leg gives $\sqrt{2x-1}.$ Conveniently, this triangle contains all of the "important" parts of our integrand, except in a non-convenient manner.

So, my question is two-fold:

(1) Does anyone see why one would be motivated to make such a substitution?

(2) Does anyone see how to extend the work involving the right triangle to get at the solution?

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This doesn't really answer your question, but (to me) making the substitution $ u = \sqrt{2x - 1}$ seems more natural. The fact that the answer turns out to be $ 2\tan^{-1}{(\sqrt{2x-1})} $ this way looks like it might suggest a motivation for considering the triangle you mentioned, if you stared at it for long enough – Daniel Freedman Jan 4 '12 at 22:06
This does not answer your question, so it is left as a comment. There is an obvious substitution $u^2=2x-1$. Then $dx=u\,du$, and after hardly any work we arrive at $\int \frac{2\,du}{1+u^2}=2\arctan u +C$. – André Nicolas Jan 4 '12 at 22:07
I'm not sure this will answer you but the side lengths of your triangle appear in the solution using this substitution: If $x={1\over 1-u}$, then $dx={ 1\over (1-u)^2}du$ and the integral becomes $$\eqalign{ \int{dx\over x\sqrt{2x-1}} &=\int {1-u\over \sqrt{2{1\over 1-u}-1 } } { 1\over( 1-u)^2}\,du\cr &=\int {1 \over \sqrt{2{1\over 1-u}-1 } } { 1\over 1-u}\,du\cr &=\int {1 \over \sqrt{{1+u\over1-u} } } { 1\over 1-u}\,du\cr &= \int {1 \over \sqrt{{1-u^2} } }\,du\cr &=\sin^{-1} u +C\cr &=\sin^{-1}{x-1\over x}+C. } $$ – David Mitra Jan 4 '12 at 22:44
I suppose I should have been more clear: for motivation I was hoping for "this substitution makes sense for this particular class of ODEs," "this integral form has a general solution of 'such and such' and, in general, we use this substitution" or even my greatest hope "this substitution is a LF/Mobius transformation and it applies to this scenario because of ..." – NoClue Jan 5 '12 at 13:53
In any event, David Mitra appears to have the closest approximation thus far regarding "inspiration" in that the argument of the arcsin contains the substitution (interestingly, one can also get a solution in arctan -- the integration constant is the medium of equality between the two solutions). – NoClue Jan 5 '12 at 13:55

Sometimes, the "motivation" for a substitution is you've solved the problem some other way and you look at the solution and you notice that there was this substitution that would have made everything work out nicely.

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This should probably be a comment, not an answer. – Bill Dubuque Jan 4 '12 at 23:07
Cut me a little slack here. The answer comes out to be $\arcsin((x-1)/x)+C$, according to David Mitra's calculations. You look at that and say, why didn't I substitute $u=(x-1)/x$ in the first place? Well, $u=(x-1)/x$ is $x=1/(1-u)$. The answer "motivates" the substitution. – Gerry Myerson Jan 5 '12 at 1:17

This won't answer the question, but it takes the geometry a bit beyond where the question left it. Consider the circle of unit radius in the Cartesian plane centered at $O=(0,1)$. Let $A=(1,0)$ and $B=(x,0)$. Let $C=(1,\sqrt{2x-1})$. Your right triangle is $ABC$, with angle $\alpha$ at vertex $B$. Another right triangle is $OAC$, with angle $\beta$ at $O$. Then $$ u=\frac{x-1}{x}= \cos\alpha=\sin(\pi/2-\alpha)=\sin\angle BCA$$ and $$ \sqrt{2x-1}=\tan\beta=\cot(\pi/2-\beta)=\cot\angle OCA. $$

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