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?
 A: 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.
A: 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 $$\int\frac{dx}{\sqrt{x^2 - a^2}}$$
A: 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
$$\int\frac{dx}{\sqrt{x^2-a^2}}=\int\frac{d(at)}{\sqrt{(at)^2-a^2}}=\int\frac{dt}{\sqrt{t^2-1}}=\cosh^{-1}t=\cosh^{-1}\frac xa.$$
A: $$\begin{align}
\text{Integrating} &\int \frac {1\cdot dx}{\sqrt {(x^2-a^2)}}\\\\
\text{Put x} &= a \cdot \cosh (\theta)\\
\Rightarrow \theta &= \cosh^{-1} \left(\frac {x}{a} \right)\\
\text{Therefore}\\ dx &= a \cdot \sinh(\theta) \cdot d\theta\\\\
\text{Now}\\
&\int \frac {a \cdot \sinh (\theta) \cdot d \theta}{\sqrt {a^2\cdot(\cosh^2(\theta)-1)}}\\\\
&\Rightarrow \frac {a \cdot \sinh (\theta) \cdot d \theta}{\sqrt {a^2\cdot \sinh ^2(\theta)}}\\
&\Rightarrow \int \frac{a \cdot \sinh (\theta) \cdot  d\theta}{a \cdot \sinh (\theta)}\\
&\Rightarrow \int 1 \cdot d\theta\\
&\Rightarrow \theta + c\\
&\Rightarrow \cosh^{-1} \left(\frac {x}{a} \right) +c
\end{align}$$
A: 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 $\dfrac{1}{\sqrt{x^2-a^2}}$
A: $\bf{My\; Solution::}$ 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 $\displaystyle I = \int\frac{dx}{y} = \int\frac{d(x+y)}{(x+y)} = \ln \left|x+y\right|+C=\ln \left|x+\sqrt{x^2-a^2}\right|+C$
A: 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 $\frac{f'}f$.)
