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Integration by substitution:

$$\int \frac {dx}{x^2-1}$$ Let $x=\sec\theta$ and $dx=\sec\theta\tan\theta \,d\theta$ $$\int \frac {dx}{x^2-1} = \int \frac{\sec\theta\tan\theta \,d\theta}{\sec^2\theta-1} = \int \frac {\sec\theta\tan\theta\, d\theta}{\tan^2\theta} = \int \frac {\sec\theta \,d\theta}{\tan\theta} $$ $$\int \frac {\sec\theta \,d\theta}{\tan\theta} = \int \frac {\cos \theta \,d\theta}{\cos\theta \sin \theta} = \int \csc\theta\, d\theta$$ $$=\ln|\csc\theta-\cot\theta|+C$$

$$=\ln| \frac{x}{\sqrt{x^2-1}}-\frac{1}{\sqrt{x^2-1}}|+C=\ln| \frac{x-1}{\sqrt{x^2-1}}|+C$$

Which is $Undefined$ for $|x|<1$


Integration by partial fractions:

$$\int \frac {dx}{x^2-1}$$ $$\int \frac {dx}{x^2-1}= \frac 12\int\frac{dx}{x-1}- \frac12\int \frac{dx}{x+1} = \frac12 \ln | x-1| - \frac12 \ln|x+1| +C$$ $$ = \frac12 \ln | \frac{x-1}{x+1}|+C$$

Which is $Defined$ for $|x|<1$ and this is right because the integrand is defined for $|x|<1$

What is the problem in the substitution method ?

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    $\begingroup$ When you declared x = $sec \theta$ you implicitly declared that |x| ≥ 1. $\endgroup$ – lulu Jul 20 '15 at 20:23
  • $\begingroup$ At the end, what did you mean by the source function? $\endgroup$ – Khallil Jul 20 '15 at 20:24
  • $\begingroup$ So, this integrand shouldn't be integrated using trigonometric substitution ? $\endgroup$ – Mohamed Mostafa Jul 20 '15 at 20:24
  • $\begingroup$ @KhallilBenyattou the integrand $\endgroup$ – Mohamed Mostafa Jul 20 '15 at 20:25
  • $\begingroup$ The integrand is only not defined for $x = \pm 1$. In any case, lulu's explained. $\endgroup$ – Khallil Jul 20 '15 at 20:28
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Let's try to clear out some of the confusion on what is going on here.

First of all, $x\mapsto \frac 1 {x^2 - 1}$ is continuous function everywhere except for $x=\pm 1$, so it is Riemann integrable on any segment not containing $\pm 1$. That said, we would very much like to find primitive function defined on $\mathbb R^2\setminus\{\pm 1\}$. Solution by partial fractions does just that, $$F_1(x) = \frac 12 \ln\left|\frac{x-1}{x+1}\right| + C$$ is defined everywhere except at $x=\pm 1$. So, if we wanted to calculate either $$I_1 = \int_{\frac 12}^{\frac 34} \frac {dx} {x^2 - 1}$$ or $$I_2 = \int_{2}^{3} \frac {dx} {x^2 - 1}$$ we can use $F_1$ with no worries.

Now, if we try to use substitution such as $x = \sec\theta$, as OP notices, we might run into some problems in the long run. The primitive function derived this way is $$ F_2(x) = \ln\left| \frac {x - 1}{\sqrt{x^2 -1}} \right| + C$$ which is defined only for $|x|>1$. Is this shocking? Well, no. As lulu points out in the comments, $|\sec\theta|\geq 1$ for any $\theta$, which is a simple consequence of the definition $\sec\theta = \frac 1 {\cos\theta}$. Thus, by substituting $x=\sec\theta$, we already gave up on $x\in\langle -1,1\rangle$, which is actually fine as long as we are trying to calculate $I_2$, but won't work for $I_1$.

So, the question is: are $F_1$ and $F_2$ both "good" solutions? More precisely, if we wanted to calculate $I_2$, can we use either of those two $F$'s?

Well, let's assume that $|x| > 1$. Then we have:

$$ \ln\left| \frac {x - 1}{\sqrt{x^2 -1}} \right| = \ln \frac {\left|x - 1\right|}{\sqrt{x^2 -1}} = \ln \frac {\sqrt{(x-1)^2}}{\sqrt{x^2 -1}} = \ln \sqrt{\frac {{(x-1)^2}}{{x^2 -1}}} = \frac 12 \ln\frac{x-1}{x+1}=\frac 12 \ln\left|\frac{x-1}{x+1}\right| $$ where the last equality holds because $x-1$ and $x+1$ have the same signs on $|x|>1$. Thus, we have shown that $F_2 = \left.F_1 \right|_{\mathbb R^2\setminus [-1,1]}$. I hope that clarifies the problem.

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  • $\begingroup$ Very well explained! $\endgroup$ – Khallil Jul 20 '15 at 21:56
  • $\begingroup$ Well. Your simplification which equated the result of the two integrations depends on the assumption of $|x|>1$ and that's fine, since for $|x|<1$ this equation can't hold. I found many problems when using trigonometric substitution in integration and don't know exactly how and why this happens.! $\endgroup$ – Mohamed Mostafa Jul 21 '15 at 1:23
  • $\begingroup$ @MohamedMostafa, do you now know what is happening here? Look at here carefully. Substitution originally works for definite integrals, and it is extended on indefinite integrals, but, one needs to be careful whether substitution will make sense on all of the domain of the integrand. In this case $x = \sec\theta$ doesn't work on whole domain of integrand, and the problem arises. $\endgroup$ – Ennar Jul 21 '15 at 10:16
  • $\begingroup$ Can we say - in short - that substitution method works if and only if the range of the substitution function is equal to the domain of the integrand to be substituted ? $\endgroup$ – Mohamed Mostafa Jul 21 '15 at 14:09
  • $\begingroup$ When you are dealing with definite integral, range of substitution function must include segment on which you are integrating (that should be obvious). When you are dealing with indefinite integral this isn't necessary, but you might end up with a function that is defined on smaller domain and then you need to find a way to extend your function to maximum domain, just as we can extend $F_1$ to $F_2$ in your exercise. You may need to differentiate the end result to make sure you really got correct antiderivative. $\endgroup$ – Ennar Jul 21 '15 at 16:20
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On simplifying

\begin{align} & \ln \frac{x-1}{\sqrt{x^2-1}}=\ln\frac{\sqrt{x-1}^2}{\sqrt{(x-1)(x+1)}} \\[6pt] = {} & \ln\frac{\sqrt{x-1}}{\sqrt{x+1}} \\[6pt] = {} & \frac 1 2 \ln \frac{x-1}{x+1} \end{align}

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    $\begingroup$ Without absolute value, what you wrote is right. but when you deal with absolute value you can't do this way. $\endgroup$ – Mohamed Mostafa Jul 20 '15 at 20:51
  • $\begingroup$ Thank you Michael for editing. Now it looks so beautiful. Please give me a link to the tutorial where i can learn to write equations and mathematical symbols on internet. $\endgroup$ – th ie Jul 20 '15 at 20:51
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    $\begingroup$ meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Mohamed Mostafa Jul 20 '15 at 20:53
  • $\begingroup$ @MohamedMostafa Thank you for the link. :) $\endgroup$ – th ie Jul 20 '15 at 20:54
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    $\begingroup$ @thie, those functions don't have same domains, i.e. these are the same functions, but only on smaller domain. $\endgroup$ – Ennar Jul 20 '15 at 21:00
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$$ \ln\frac{|x-1|}{\sqrt{|x^2-1|}} = \ln\frac{\sqrt{|x-1|}\sqrt{|x-1|}}{\sqrt{|x-1|}\sqrt{|x+1|}} = \ln\sqrt{\frac{|x-1|}{|x+1|}} = \frac 1 2 \ln\left|\frac{x-1}{x+1}\right| $$

In response to comments I've made this more complete than it was. Notice that

  • I take no square roots of anything except non-negative numbers; and
  • $|AB| = |A||B|$, so $|x^2-1|=|x-1||x+1|$; and
  • $\sqrt{AB} =\sqrt A \sqrt B$ if $A\ge0$ and $B\ge0\vphantom{\dfrac 1 1}$, so the separation into two square roots is valid; and
  • $\sqrt A/\sqrt B = \sqrt{A/B\,{}}$, if $A\ge0$ and $B\ge0$, so the second equality is valid; and
  • $|A|/|B| = |A/B|$, so the last equality is valid.

PS in response to comments: The problem with the trigonometric substitution is only that it is valid only when $|x|>1$, since $\sec\theta\ge 1$ for all values of $\theta$ and those points where $|x|=1$ are not in the domain.

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  • $\begingroup$ Without absolute value, what you wrote is right. but when you deal with absolute value you can't do this way. $\endgroup$ – Mohamed Mostafa Jul 20 '15 at 20:52
  • $\begingroup$ Correct .... This is not a complete answer. It works when $x>1$. But with this information, the original poster should be able to figure everything out. $\endgroup$ – Michael Hardy Jul 20 '15 at 20:55
  • $\begingroup$ Exactly. The problem is Rogawski - the author of the text I study - did the same simplification and I don't know if it was a mistake! $\endgroup$ – Mohamed Mostafa Jul 20 '15 at 20:58
  • $\begingroup$ You most left hand side is NOT the result of integration by substitution. Then, this equation due to simplification can't hold for $|x|<1$ and the problem still there. $\endgroup$ – Mohamed Mostafa Jul 21 '15 at 1:25

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