# 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.

• By "faster," do you mean "analytically," as opposed to "numerically"? Sep 21 '15 at 14:55
• analytically ofc. Sep 21 '15 at 14:59
• what do you mean 'euler' substitution? Sep 21 '15 at 15:00
• $x=\cosh(t)$ works pretty nice. Sep 21 '15 at 15:06
• Hyperbolic substitution must work nice, we're integrating $x^2-y^2=1$. Sep 21 '15 at 15:09

## 3 Answers

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 $$\boxed{\color{blue}{\int\sqrt{x^2-1}\,dx=\frac{1}{2}x\sqrt{x^2-1}-\frac{1}{2}\ln |x+\sqrt{x^2-1}|+C}}$$

• How are you going from $\sqrt{\tan^2 t}$ to $\tan t$, given that $t=\sec^{-1} x$, and the range of $\sec^{-1} x$ is typically $[0, \pi]$, so that $\tan t$ could be negative? Shouldn't it be $\sqrt{\tan^2 t}=|\tan t|$? Jan 18 '21 at 10:21

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 $$\frac{1}{2}\int (\cosh{2t}-1) \, dt = \frac{1}{4}\sinh{2t}-\frac{t}{2}+C = \frac{1}{2}(\cosh{t}\sqrt{\cosh^2{t}-1}-t)+C = \frac{1}{2}(x\sqrt{x^2-1}-\arg\cosh{x})+C$$

Integration without substitution

Notice, there is another method without any substitution, using integration by parts, let
$$I=\int\sqrt{x^2-1}\ dx$$$$=\int \sqrt{x^2-1}\cdot 1dx$$ $$I=\sqrt{x^2-1}\int 1\ dx-\int \left(\frac{d}{dx}\left(\sqrt{x^2-1}\right)\cdot \int1 dx\right)dx$$ $$I=\sqrt{x^2-1}(x)-\int \frac{x(2x)}{2\sqrt{x^2-1}}dx$$ $$I=x\sqrt{x^2-1}-\int \frac{(x^2-1)+1}{\sqrt{x^2-1}}dx$$ $$I=x\sqrt{x^2-1}-\int\sqrt{x^2-1} dx+\int \frac{1}{\sqrt{x^2-1}}dx$$

$$I=x\sqrt{x^2-1}-I+\ln|x+\sqrt{x^2-1}|+C$$ $$2I=x\sqrt{x^2-1}+\ln|x+\sqrt{x^2-1}|+C$$ $$I=\color{red}{\frac{1}{2}\left(x\sqrt{x^2-1}+\ln|x+\sqrt{x^2-1}|\right)+C}$$

• Though your answer is clean, OP specifically wanted a trigonometric substitution Sep 21 '15 at 15:09
• The OP supposed that there is a trig substitution to do the job. But looking at the OP's trials I would assume that he may be open to alternative ,methods as well, as we all should... So time for an upvote to compensate the downvote. Sep 21 '15 at 15:30