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I see people use the substitution:

$$u = \frac{1 − x}{1 + x}$$

in integrals, but I have no idea when to use it. I can see that the function

$$f(x) = \frac{1 − x}{1 + x},$$

has an interesting property when trying to find the inverse function:

$$f(f(x)) = x.$$

I want to know when one should use this substitution and why, for example when one sees $\sqrt{1 − x^2}$, the first thing that comes to mind is $x = \sin u$ because $1 − \sin^2 u = \cos^2 u$, when one sees $1 + x^2$, the first thing that comes to the is $x = \tan u $ because $1 + \tan^2 u = \sec^2 u$. In these examples, the substitution is used to make use of some formula that will simplify the equation in the integral, but I don’t see the reason why people use:

$$u = \frac{1 − x}{1 + x},$$

or

$$u^2 = \frac{1 − x}{1 + x}.$$

Another question is: Does this substitution have a name?

Alwaz
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  • "I want to know when one should use this substitution" $\int_0^1f\left(\frac{1-x}{1+x}\right)dx$ seems a good candidate. I don't know of it does have a name. – Zima Jun 03 '24 at 07:51
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    See if this answers your question. – Zacky Jun 03 '24 at 08:51
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    Note that this transformation is a specific case of the more general transformation $x = \frac{at , + , b}{t , + , 1},$ which is used in Section 5 of this MSE answer for a nontrivial choice of the constants $a$ and $b,$ and one purpose of this transformation is explained in Section 4 of that answer. – Dave L. Renfro Jun 03 '24 at 11:12
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    We call substitutions like these “Linear fractional transformations”: https://mathworld.wolfram.com/LinearFractionalTransformation.html – David H Jun 03 '24 at 21:27
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    As a fun exercise, you can use the substitution $u = \frac{1-x}{1+x}$ on integrals of the form $\int_{0}^{1}\frac{\arctan x}{bx^{2}-bx+x+1}dx$ where they are Riemann integrable for $-\infty<b<3+2\sqrt{2}$. – Accelerator Jun 07 '24 at 16:43

1 Answers1

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Yes actually it comes from a specific source :

Namely :

First

Having $x\to \sqrt{1-x^2}$, you want to get rid of the square root because it isn't convenient for integration. And a way to make it disappears is setting $x=\cos(\theta)$ because $1-\cos(\theta)^2=\sin(\theta)^2$, take the square root and then you have the $|\sin(\theta)|$, no more square root !

So $$\int_0^1 \sqrt{1-x^2}= \int_{0}^{\pi/2}\sin(\theta)\cos(\theta)d\theta$$

Secondly

You ends up with a trigonometric integral.

For trigonometric integrand you can use often the following substitution :

$$ x= \tan(\theta/2) $$ with $\theta$ a new variable.

Why ?

Because it allows to get from trigonometric to polynomial (without roots) $$ \cos(\theta) =\dfrac{1-x^2}{1+x^2} \ \ \text{See how it it close to your problem}$$

and

$$ \sin(\theta)= \dfrac{2x}{1+x^2} $$

and also (!)

$$ \tan(\theta)= \dfrac{2x}{1-x^2} $$

It allow you to turn any trigonometric integral into polynomial (according you get rid of the roots if more complex roots occur)

Is it better for you ?

EDX
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