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$$\int{\frac{1}{a^2-x^2}}dx\tag{1}$$

When finding the above anti-derivative, $x$ is substituted with $a\sin(\theta)$. However, the range of $x$ is $\mathbb{R}-\{-a,a\}$ while the range of $a\sin(\theta)$ is $[-a,a]$.

Graph of $\frac{1}{a^2-x^2}[a=0.5]$

Needless to say,

$$\mathbb{R}-\{-a,a\}\ne[-a,a]$$

So, how can $a\sin\theta$ be substituted for $x$ in $(1)$ when the range of $x$ and $a\sin\theta$ are not the same?

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tryingtobeastoic
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    The domain of the integrand is $\mathbb R- {-a,a}$. I don't see a problem. The substitution can be made when $x\in(-a,a)$. Otherwise, we can substitute$x=a\csc u$, or $x=a\sec u$. – Laxmi Narayan Bhandari Jan 27 '22 at 08:57

2 Answers2

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$$\int{\frac{1}{a^2-x^2}}dx\tag{1}$$

the range of $x$ is $\mathbb{R}-\{-a,a\}$ while the range of $a\sin(\theta)$ is $[-a,a]$

how can $a\sin\theta$ be substituted for $x$ in $(1)$ when the range of $x$ and $a\sin\theta$ are not the same?

On the integration domain $(-a,a)$ (let's call this interval $A$), the substitution $x=a\sin\theta$ with range $[-a,a]$ (let's call this interval $B$) is valid because $A\subseteq B.$

On the separate integration domains $(-\infty,-a)$ and $(a, \infty)$ (let's call these intervals $C$ and $D,$ respectively), the substitution $x=a\sec\theta$ with range $[-\infty,-a]\cup[a,\infty]$ (let's call this set $E$) is valid because $C\subseteq E$ and $D\subseteq E.$

(Alternatively, use the hyperbolic substitutions $x=-a\cosh\theta$ and $x=a\cosh\theta$ for integration domains $C$ and $D,$ respectively.)

But this integral is easiest found using partial fractions, without any substitution.

ryang
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  • Hmm, I see. However, in the case of $\int{\frac{1}{x^2-a^2}}\mathbb{dx}$, $a\sec\theta$ can be substituted for $x$ without any problem as the range of $a\sec\theta=\mathbb{R}-{a,-a}=$ range of $x$: we don't need any other trig functions other than $a\sec\theta$ in this case like we needed in the case of $\int{\frac{1}{a^2-x^2}}\mathbb{dx}$, right? – tryingtobeastoic Jan 27 '22 at 11:34
  • @tryingtobeastoic 1. As Laxmi suggested, $\csc$ too works, but of course we would rather deal with $\sec$ (& $\tan)$ than $\csc$ (& $\cot).$ In fact, in this answer, I explained why I'd rather deal with a $\cosh$ sub than a $\sec$ sub. 2. A valid substitution is not necessarily a useful substitution. 3. More importantly, an integrand like your suggested one requires no substitution (on the other hand, we might consider a substitution if one of the forms in the aforementioned answer appears in the integrand). – ryang Jan 27 '22 at 11:51
  • @tryingtobeastoic I) $a\sec\theta$ does not have range $\mathbb{R}-{a,-a}$; read my 2nd paragraph carefully. II) In fact, please just re-read my entire answer slowly; then consider whether paragraph 1 applies equally whether the integrand is $\frac{1}{x^2-a^2}$ or $\frac{1}{a^2-x^2}.$ – ryang Jan 27 '22 at 13:58
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When $x\in (a, +\infty)$, you can substitute $x = a \cosh \theta$, thus \begin{equation} \int \frac{d x}{a^2 - x^2} = \int\frac{1}{a^2}\frac{a\sinh \theta d\theta}{1 -\cosh^2 \theta} = -\frac{1}{a}\int\frac{d\theta}{\sinh\theta} \end{equation} You can then use $u = e^\theta$ or $t = \tanh\frac{\theta}{2}$.

Of course, you could decompose directly from the beginning \begin{equation} \frac{1}{a^2-x^2} = \frac{1}{2a}\left(\frac{1}{a-x} + \frac{1}{a+x}\right) \end{equation}

Gribouillis
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  • Someone downvoted faster than their own shadow! I wonder why. Using a hyperbolic trigonometry substitution is quite standard when normal circular trigonometry does not work. – Gribouillis Jan 27 '22 at 10:27