Integrating $\frac 1 {\sqrt{x^2 - a^2}}$

I'm trying to understand how to integrate: $$\frac 1 {\sqrt{x^2 - a^2}}$$ I tried substituting $t$ as $\sqrt{x^2 - a^2}$, but I can't get an answer that way, and I don't know any other way to integrate other than by parts, which I don't think can be used here. How should I integrate this?

• Let $x=a\sec(t)$.
– user121880
Commented Aug 5, 2014 at 15:49
• For a magic substitution, let $u=x+\sqrt{x^2-a^2}$. Commented Aug 5, 2014 at 16:14
• @AndréNicolas I've got the correct antiderivative using the method in the answers.. but I'm stuck with this magic substitution. Commented Aug 5, 2014 at 16:22
• We have $du=(1+\frac{x}{\sqrt{x^2-a^2}})\,dx=\frac{x+\sqrt{x^2-a^2}}{\sqrt{x^2-a^2}}\,dx$. So $\frac{du}{u}=\frac{1}{\sqrt{x^2-a^2}}\,dx$. Integrate. We get $\ln u+C$. Commented Aug 5, 2014 at 16:27
• Brilliant! That's very interesting. Commented Aug 5, 2014 at 16:32

There is a difference between $$\int\frac{dx}{\sqrt{x^2-a^2}} \text{ and }\int\frac{da}{\sqrt{x^2-a^2}}.$$ Of course it is conventional to assume the former is meant, but writing $dx$ or $da$ makes that explicit.

Both of them beg for trigonometric substitutions.

Assuming $dx$ and not $da$, then \begin{align} x^2-a^2 & = (a\sec\theta)^2 -a^2 = a^2(\sec^2\theta-1)=a^2\tan^2\theta \\[8pt] a^2-x^2 & = a^2 - (a\sin\theta)^2 = a^2(1-\sin^2\theta)=a^2\cos^2\theta \\[8pt] a^2+x^2 & = a^2 + (a\tan\theta)^2 = a^2(1+\tan^2\theta)=a^2\sec^2\theta \end{align} (The third one above should be used regardless of whether it's $dx$ or $da$, but interchanging $dx$ and $da$ corresponds to interchanging the roles of the first two equalities above.)

All three of these use Pythagorean identities from trigonometry.

• How can we substitute $asec\theta$ for x? Commented Aug 5, 2014 at 16:00
• Note that it is also possible to use hyperbolic functions for the first of these substitutions because $\cosh^2 x-\sinh^2x =1$ has the same form as $\sec^2 x-\tan^2 x =1$ Commented Aug 5, 2014 at 16:01
• @ABajaj : In the expression $\displaystyle\int\frac{dx}{\sqrt{x^2-a^2}}$, the variable $x^2$ necessarily takes values $\ge a^2$ (unless we bring in complex numbers, which opens a can of worms that is not mentioned here). So $|x/a|\ge1$, and thus there is some value of $\theta$ for which $x/a = \sec\theta$. If you have an objection to that, can you say what it is? ${}\qquad{}$ Commented Aug 5, 2014 at 16:09

Hint: Use the substitution $\underbrace{x = a\sec \theta}_{\Large \theta = \operatorname{arcsec}\left(\frac xa\right)}\implies dx = a\sec\theta \tan\theta$

And recall that $\tan^2 \theta = \sec^2 \theta - 1$.

See this link for an outline of dealing with integrals containing $x^2 - a^2$ where the variable with respect to which we are integrating is $x$ (i.e., where $dx$ accompanies your function under the integral sign: such as $$\int\frac{dx}{\sqrt{x^2 - a^2}}$$

\begin{align} \text{Integrating} &\int \frac {1\cdot dx}{\sqrt {(x^2-a^2)}}\\\\ \text{Put x} &= a \cdot \cosh (\theta)\\ \Rightarrow \theta &= \cosh^{-1} \left(\frac {x}{a} \right)\\ \text{Therefore}\\ dx &= a \cdot \sinh(\theta) \cdot d\theta\\\\ \text{Now}\\ &\int \frac {a \cdot \sinh (\theta) \cdot d \theta}{\sqrt {a^2\cdot(\cosh^2(\theta)-1)}}\\\\ &\Rightarrow \frac {a \cdot \sinh (\theta) \cdot d \theta}{\sqrt {a^2\cdot \sinh ^2(\theta)}}\\ &\Rightarrow \int \frac{a \cdot \sinh (\theta) \cdot d\theta}{a \cdot \sinh (\theta)}\\ &\Rightarrow \int 1 \cdot d\theta\\ &\Rightarrow \theta + c\\ &\Rightarrow \cosh^{-1} \left(\frac {x}{a} \right) +c \end{align}

From a table of derivatives, $\frac{d}{dx}\left[\cosh^{-1}x\right]=\frac1{\sqrt{x^2-1}}$.

And by a simple change of scale $$\int\frac{dx}{\sqrt{x^2-a^2}}=\int\frac{d(at)}{\sqrt{(at)^2-a^2}}=\int\frac{dt}{\sqrt{t^2-1}}=\cosh^{-1}t=\cosh^{-1}\frac xa.$$

• How is this answer the same as the other one? Commented Aug 5, 2014 at 16:52
• You can start from $x=\cosh y=\frac{e^y+e^{-y}}2$ and invert it to get the formula for $\cosh^{-1}$.
– user65203
Commented Aug 5, 2014 at 16:58

May be we can remark that $\cosh(\text{argcosh}(x))=x$ and so by derivation, $$\text{argcosh}'(x)=\frac{1}{\sinh(\text{argcos}(x))}=\frac{1}{\sqrt{x^2-1}}$$

I let you generalise the case where we have $\dfrac{1}{\sqrt{x^2-a^2}}$

$\bf{My\; Solution::}$ Let $\displaystyle I = \int\frac{1}{\sqrt{x^2-a^2}}dx$

Let $\left(x^2-a^2\right)=y^2\;,$ Then $\displaystyle 2xdx = 2ydy\Rightarrow \frac{dx}{y}=\frac{dy}{x}$

Now Using Ration and Proportion, We Get $\displaystyle \frac{dx}{y}=\frac{dy}{x}=\frac{d(x+y)}{(x+y)}$

So Integral $\displaystyle I = \int\frac{dx}{y} = \int\frac{d(x+y)}{(x+y)} = \ln \left|x+y\right|+C=\ln \left|x+\sqrt{x^2-a^2}\right|+C$

• This is an interesting solution, but using ratios and proportions here seems a bit odd? Commented Aug 20, 2014 at 14:33

Here is the most far-fetched solution I can think of $$\int\frac{dx}{\sqrt{x^2-a^2}}=\int\frac1{\sqrt{x^2-a^2}}\frac{\sqrt{x^2-a^2}+x}{x+\sqrt{x^2-a^2}}dx=\int\frac{1+\frac x{\sqrt{x^2-a^2}}}{x+\sqrt{x^2-a^2}}dx=\ln(\sqrt{x^2-a^2}+x).$$ (The last integrand is of the form $\frac{f'}f$.)

• Why do you call this far fetched? Commented Aug 5, 2014 at 16:33
• Because nothing tells you that introducing the expression $\sqrt{x^2-a^2}+x$ will yield such a nice simplification. A rabbit pulled out of a hat.
– user65203
Commented Aug 5, 2014 at 16:36
• So... how did you think of it? (It's the same suggestion Andre Nicolas suggested) Commented Aug 5, 2014 at 16:37
• Just by cheating (looking at the solution). Anyway, I also recognized the derivative of the $argch$ function (a close cousin of $arcsin$) for which the closed formula is known.
– user65203
Commented Aug 5, 2014 at 16:38
• I've been searching, but I haven't found anything about argch. What is it? Commented Aug 5, 2014 at 16:48