Let $a>0.$ Using substitution, determine $$\int{\frac{1}{x^2-a^2}}\,\mathbb{d}x.$$

My book's attempt:

$$\int{\frac{1}{x^2-a^2}}\mathbb{d}x\\ [\text{let}\ x=a\sec\theta,\ \therefore dx=a\sec\theta\tan\theta d\theta]\\ =\int{\frac{1}{a^2\sec^2\theta-a^2}}a\sec^2\theta d\theta\\ =\int{\frac{1}{a^2(\sec^2\theta-1)}}a\sec\theta\tan\theta d\theta\\ =\int{\frac{1}{a^2\tan^2\theta}}a\sec\theta\tan\theta d\theta\\ =\frac{1}{a}\int{\frac{\sec\theta}{\tan\theta}}d\theta\\ =\frac{1}{a}\int{\frac{\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}}}d\theta\\ =\frac{1}{a}\int{\csc\theta d\theta}\\ =\frac{1}{a}\int{\frac{\csc\theta(\csc\theta+\cot\theta)}{(\csc\theta+\cot\theta)} d\theta}\\ =-\frac{1}{a}\int{-\frac{\csc^2\theta+\csc\theta\cot\theta}{(\csc\theta+\cot\theta)} d\theta}\\ =-\frac{1}{a}\ln|\csc\theta+\cot\theta|+C$$

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$$=-\frac{1}{a}\ln\left|\frac{x}{\sqrt{x^2-a^2}}+\frac{a}{\sqrt{x^2-a^2}}\right|+C\\ =-\frac{1}{a}\ln\left|\frac{x+a}{\sqrt{x-a}\sqrt{x+a}}\right|+C\\ =-\frac{1}{a}\ln\left|\frac{\sqrt{x+a}}{\sqrt{x-a}}\right|+C\\ =-\frac{1}{a}{\ln\left|\frac{x+a}{x-a}\right|}^{\frac{1}{2}}+C\\ =-\frac{1}{2}.\frac{1}{a}\ln\left|\frac{x+a}{x-a}\right|+C\\ =\frac{1}{2a}{\ln(\left|\frac{x+a}{x-a}\right|)}^{-1}+C\\ =\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C.$$

Someone told me that this work contains many errors. The second line clearly has a typo ($\sec^2\theta$ should be replaced with $\sec\theta\tan\theta$ there); what are the other errors?

Related: 1, 2

  • 5
    $\begingroup$ Other than the error that you found, which is magically corrected in the next line, the rest looks fine to me. However, partial fractions would be so much easier. There is a question of whether the substitution $x = a\sec \theta$ is valid in the case that $|x| < 1.$ It is not if we insist $\theta$ be a real number. $\endgroup$
    – Doug M
    Jan 28 at 7:16
  • $\begingroup$ which book is this? $\endgroup$ Jan 28 at 8:38
  • 1
    $\begingroup$ @lalittolani it's a relatively unknown Bangladeshi book $\endgroup$ Jan 28 at 8:40
  • 2
    $\begingroup$ The author's unspecified assumption is that the integration interval is $(a,\infty).$ (Why do I say this?)$\quad$ Even though the final answer is indeed also true for the intervals $(-\infty,-a)$ and $(-a,a),$ the author has not proven this.$\quad$ A good learning exercise is to repair/complete the author's proof (still using substitution, even though partial fractions is really the way to go here). With your work, perhaps edit your Question or self-Answer it below. $\endgroup$
    – ryang
    Jan 28 at 12:34

1 Answer 1

  1. The given integral $\displaystyle\int{\frac{1}{x^2-a^2}}\,\mathbb{d}x$ properly exists on each of three possible intervals: $$(-\infty,-a),(-a,a)\;\text{and}\;(a, \infty).$$

  2. The substitution $$x=a\sec\theta$$ is inapplicable for $x\in(-a,a),$ so the author's proof is invalid for this interval.

  3. Since the author writes $$\sqrt{x^2-a^2} = \sqrt{x-a}\sqrt{x+a},$$ their proof is technically invalid for $x\in(-\infty,-a)$ too.

    Here's a correction so that their proof is at least valid for $(-\infty,-a)\cup(a, \infty):$ $$-\frac1a\ln\left|\frac{x}{\sqrt{x^2-a^2}}+\frac{a}{\sqrt{x^2-a^2}}\right|+C\\ =-\frac1a\ln\frac{|x+a|}{\sqrt{|x+a||x-a|}}+C\\ =-\frac1a\ln\frac{\sqrt{x+a}}{\sqrt{x-a}}+C\\ =\frac1{2a}\ln\left|\frac{x-a}{x+a}\right|+C.$$

  4. Notice that the author writes $$\operatorname{cosec}\theta=\frac x{\sqrt{x^2-a^2}}\\ \cot\theta=\frac a{\sqrt{x^2-a^2}}.$$ This is true for $x\in(-\infty,-a)\cup(a, \infty)$ if the implicit substitution $x=a\sec\theta$ has domain $$\theta\in\left(0,\frac\pi2\right)\cup\left(\pi,\frac{3\pi}2\right),$$ which is better explicitly specified than left tacit.

    If the substitution has the more conventional domain $$\theta\in\left(0,\frac\pi2\right)\cup\left(\frac\pi2,\pi\right),$$ then $$\operatorname{cosec}\theta=\frac {|x|}{\sqrt{x^2-a^2}}\\ \cot\theta=-\frac {ax}{|x|\sqrt{x^2-a^2}},$$ and the subsequent algebra is to be modified accordingly.


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