Indefinite integral of $\int\sqrt{x^2-1}dx$ How to calculate $\int\sqrt{{x^2}-1}dx$?
I tried to use 1st and 3rd Euler substitution and got the answer, but I suppose there's an trigonometrical replacing that gives the answer analytically.
 A: Take $x=\sec t$, then $dx=\sec t\tan t\, dt$ and
\begin{align}
\int\sqrt{x^2-1}\,dx&=\int\sqrt{\sec^2 t -1}\sec t\tan t\,dt\\
&=\int\sqrt{\tan^2 t}\sec t\tan t\,dt\\
&=\int\sec t\tan^2 t\,dt
\end{align}
Next, integration by parts can be applied:
\begin{align}
\int\sec t\tan^2 t\,dt&=\tan t\sec t-\int\sec^3 t\,dt\\
&=\tan t\sec t-\int(\tan^2 t+1)\sec t\,dt\\
&=\tan t\sec t-\int \sec t\tan^2 t\,dt -\int\sec t\, dt\\
2\int\sec t\tan^2 t\,dt&=\tan t\sec t-\ln |\sec t+\tan t|\\
\int\sec t\tan^2 t\,dt&=\frac{1}{2}\tan t\sec t-\frac{1}{2}\ln |\sec t+\tan t|+C
\end{align}
Then, since $\sec t=x$ we can put $\tan t=\sqrt{\sec^2 t-1}=\sqrt{x^2-1}$, hence
$$\boxed{\color{blue}{\int\sqrt{x^2-1}\,dx=\frac{1}{2}x\sqrt{x^2-1}-\frac{1}{2}\ln |x+\sqrt{x^2-1}|+C}}$$
A: As an alternative, take $x=\cosh{t}$. Then $dx=\sinh{t} \, dt$ and
$$ \int \sqrt{x^2-1} \, dx = \int \sqrt{\cosh^2{t}-1} \sinh{t} \, dt = \int \sinh^2{t} \, dt. $$
Now,
$$ \cosh{2t} = \cosh^{2}{t}+\sinh^2{t} = 1+2\sinh^2{t}, $$
so this is
$$ \frac{1}{2}\int (\cosh{2t}-1) \, dt = \frac{1}{4}\sinh{2t}-\frac{t}{2}+C = \frac{1}{2}(\cosh{t}\sqrt{\cosh^2{t}-1}-t)+C = \frac{1}{2}(x\sqrt{x^2-1}-\arg\cosh{x})+C $$
A: Integration without substitution
Notice, there is another method without any substitution, using integration by parts, let
$$I=\int\sqrt{x^2-1}\ dx$$$$=\int \sqrt{x^2-1}\cdot 1dx$$
$$I=\sqrt{x^2-1}\int 1\ dx-\int \left(\frac{d}{dx}\left(\sqrt{x^2-1}\right)\cdot \int1 dx\right)dx$$
$$I=\sqrt{x^2-1}(x)-\int \frac{x(2x)}{2\sqrt{x^2-1}}dx$$
$$I=x\sqrt{x^2-1}-\int \frac{(x^2-1)+1}{\sqrt{x^2-1}}dx$$
$$I=x\sqrt{x^2-1}-\int\sqrt{x^2-1} dx+\int \frac{1}{\sqrt{x^2-1}}dx$$
$$I=x\sqrt{x^2-1}-I+\ln|x+\sqrt{x^2-1}|+C$$
$$2I=x\sqrt{x^2-1}+\ln|x+\sqrt{x^2-1}|+C$$
$$I=\color{red}{\frac{1}{2}\left(x\sqrt{x^2-1}+\ln|x+\sqrt{x^2-1}|\right)+C}$$
