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How can I solve the integral

$$\int \frac{1}{\sqrt{x^2-1}}dx$$ without using any trigonometric substitution? I know that $$\frac{d}{dx} \text{arccosh} (x)=±\frac{1}{\sqrt{x^2-1}}$$

Does the fundamental theorem of calculus apply? If so, I could conclude, for $x>0$, the integral as $$\text{arccosh}(x)+C$$

Travis Willse
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mavavilj
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  • Yes, but this isn't "solving the integral" so much as it is verifying a solution. – Travis Willse Mar 29 '16 at 12:42
  • Looking at the WA solution I see a different sign for $x < -1$ (third plot), so it might need a deeper look at the definitions. – mvw Mar 29 '16 at 12:42
  • Of course, this antiderivative can be derived by applying the reverse substitution $x = \cosh u$, $dx = \sinh u$, which gives $\int ,du$, but at least from a complex point of view this is trigonometric. One can obscure this fact by instead working with the exponential form of these functions and substitute $x = \tfrac{1}{2}(e^u - e^{-u})$. – Travis Willse Mar 29 '16 at 12:49
  • $$\int\frac{\mathrm dx}{\sqrt{x^2-1}}=\text{sgn}x\cosh^{-1}|x|+C$$ – Integreek Nov 28 '24 at 15:27

5 Answers5

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I don't see why you won't do a trigonometric sub, (put $x=\sec \theta$) , but anyway, this integral can be written as: $$\int \frac{dx}{x\sqrt{1-\frac{1}{x^2}}}$$ Substitute $1-\frac{1}{x^2}=t^2$,

You get $$\frac{2}{x^3}dx=2tdt$$ $$\implies \frac{dx}{x}.\frac{1}{x^2}=tdt$$ $$\implies \frac{dx}{x}.(t^2-1)=-tdt$$ $$\implies \frac{dx}{x}=\frac{-tdt}{t^2-1}$$ Now, put everything back in the integral to get: $$\int \frac{-dt}{t^2-1}$$ Which may be done by partial fractions or by splitting the numerator as $\frac{1}{2} (t+1-(t-1))$

Nikunj
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With $y= \sqrt{x^2-1}$ and $y\ dy=x\ dx$ \begin{align} \int \frac1{\sqrt{x^2-1}}dx =& \int \frac{dx}y= \int \frac{(1+\frac xy)dx}{x+y} =\int \frac{dx+dy}{x+y}\\ =&\ \ln|x+y|=\ln \left|x+\overset{}{\sqrt{x^2-1}}\right|+C \end{align}

Quanto
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Applying the rationalizing First Euler Substitution, $$\sqrt{x^2 - 1} = -x + t, \qquad dx = \frac{t^2 + 1}{2 t} \,dt,$$ and simplifying leaves $$\int \frac{dt}{t} = \log |t| + C = \log \left\vert\sqrt{x^2 - 1} + x\right\vert + C = \operatorname{arcosh} x + C.$$

Travis Willse
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Let $x^2-1=u^2$ thus $dx=\frac{udu}{x}$ but $ x=\sqrt{u^2+1}$ thus integral becomes after simplification $$\int \frac{1}{\sqrt{u^2+1}}du=ln(u+\sqrt{u^2+1^2})+C$$

Archis Welankar
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  • So, you know $\int\dfrac{du}{\sqrt{u^2+1}}$ but not $\int\dfrac{du}{\sqrt{u^2-1}}$ – lab bhattacharjee Mar 29 '16 at 12:48
  • This begs the question in a way, as the latter integral is apparently no easier to handle without trigonometric substitution than the former. – Travis Willse Mar 29 '16 at 12:50
  • Yes its given in my book under standard formulae why isnt this known?? – Archis Welankar Mar 29 '16 at 12:50
  • @ArchisWelankar I think it is not given in some books. So in case anyone is wondering, it makes use of Euler substitution. https://en.wikipedia.org/wiki/Euler_substitution – lEm Mar 29 '16 at 13:06
  • @ArchisWelankar how do you integrate that directly? In my book the result of the question that the OP asked is given directly, should I write it? – Nikunj Mar 30 '16 at 10:31
  • No op said he knows the result just wanted another way if there was for integration – Archis Welankar Mar 30 '16 at 10:39
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The function $\operatorname{arcosh}$ is only defined over $[1,\infty)$ where it satisfies $x=\cosh\operatorname{arcosh}x$, so $$ 1=\sinh\operatorname{arcosh}x\operatorname{arcosh}'x $$ which means $$ \operatorname{arcosh}'x=\frac{1}{\sin\operatorname{arcosh}x}= \frac{1}{\sqrt{x^2-1}} $$ for $x>1$ (it is not differentiable at $1$).

Thus the most general antiderivative of your function is $$ \begin{cases} \operatorname{arcosh}x+c_+ & \text{for $x>1$} \\[6px] -\operatorname{arcosh}(-x)+c_- & \text{for $x<-1$} \end{cases} $$ where $c_+$ and $c_-$ are arbitrary constants.

There's no need to do substitutions, trigonometric or not, if you already know about that function.

egreg
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