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For evaluating $\int \dfrac{dx}{x^2 - a^2}$, how can we make the substitution $x= a\sec \theta $ because $\sec \theta$ can be 1 and then that would give 1/0 form.

So how can we do that and why does it work? Why not use $a\tan \theta$?

And: $a^2 \sec^2 \theta$ misses the values less than $a^2$. What do we do about that?

Archer
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    $x$ might be equal to $a$ or $-a$, too. You don’t appear to be troubled by that. – amd Jul 04 '18 at 07:45
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    You can make any substitution you want. Some of them work better than others. But, your main worry seems to be about whether a substitution makes sense for the interval where you want to use the resulting indefinite integral. Asking that question alone would earn points in my exams! So $\large{+1}$. – Jyrki Lahtonen Jul 04 '18 at 07:47
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    Why dont you use partial fraction decomposition? – Tony Ma Jul 04 '18 at 07:48
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    The answer is the usual (when looking for indefinite integrals). When you get a candidate for a primitive, by whatever means, possibly restricted only to a small interval, you can always verify it by differentiating. Usually straightforward because differentiating elementary functions is a rather mechanical process. Then you can use that as a justification. Do check that you won't violate domains of definition :-) – Jyrki Lahtonen Jul 04 '18 at 07:50
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    Anyway, it is great to think of different substitutions. When dealing with definite integrals you need to pay attention to that (when finding the limits w.r.t. the new variable). But, here I would just use partial fractions :-) – Jyrki Lahtonen Jul 04 '18 at 07:52

2 Answers2

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An option:

$\dfrac{1}{ x^2-a^2}=\dfrac{1}{(x-a)(x+a)}=$

$\dfrac{1}{2a}[ \dfrac{1}{x-a} -\dfrac{1}{x+a}]$.

Peter Szilas
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0

Another hint:

$$(\operatorname{argtanh})'(x)=\dfrac1{1-x^2},$$ so you can obtain $\;\displaystyle\int\frac{\mathrm dx}{a^2-x^2}$ with the substitution $\; x=at$. Also, we know that $$\operatorname{argtanh}x=\frac 12\ln\biggl(\frac{1+x}{1-x}\biggr).$$

Bernard
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