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I was looking at some integrals to do with trigonometric substitutions and I stumbled across this one

$$\int\frac{1}{\sqrt{x^2-1}}dx$$

I know you can do it with a regular trigonometric substitution or just use a hyperbolic substitution but I was wondering if you can do it the following way.

$$ \int \frac{1}{\sqrt{x^2-1}} dx = \int \frac{\cos \theta}{\sqrt{-\cos^2\theta}}d\theta = \int \frac{1}{i}d\theta = \frac{1}{i}\arcsin x, $$ where the $x=\sin \theta$ substitution was used. Could anybody please explain to me why I don't get the same result as one would get if a hyperbolic or other trigonometric substitution was used?

Thanks in advance.

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  • $\begingroup$ $$\int\frac{1}{\sqrt{x^2-1}}dx=\text{arcosh}(x)+C=\pm i\arccos(x)+C$$ $$=\pm i\left(\frac\pi2-\arcsin(x)\right)+C=\mp i\arcsin(x)+\left(C\pm i\frac\pi2\right)$$ $$=\pm\frac1i\arcsin(x)+D$$ $\endgroup$
    – mr_e_man
    Commented Feb 13, 2020 at 22:55
  • $\begingroup$ @mr_e_man thanks for the reposnse. Could you please clarify or link the source for where $\text{arccosh (x)} = +/- \arccos (x)$? Thanks in advance! $\endgroup$ Commented Feb 13, 2020 at 23:45
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    $\begingroup$ $$\cosh(\pm i\theta)=\sum_{k=0}^\infty\frac{(\pm i\theta)^{2k}}{(2k)!}=\sum_{k=0}^\infty(-1)^k\frac{\theta^{2k}}{(2k)!}=\cos(\theta)$$ $\endgroup$
    – mr_e_man
    Commented Feb 13, 2020 at 23:48

1 Answer 1

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To keep you original integral real you require $$ x^2 \gt 1 $$ If you let $$ x=\sin \theta$$ then for real $\theta$ it must be true that $$ -1 \le x \le 1 \Rightarrow x^2 \le 1 $$

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  • $\begingroup$ Hi, thanks for your answer. I have a question. Why is $x^2>1$ required? Why can't you consider the case for $x^2<1$ and then consider the real part of the integral? Thanks in advance. $\endgroup$ Commented Feb 13, 2020 at 21:33
  • $\begingroup$ There are multiple things to think about if integrating a complex valued function. You need to consider carefully the region your function is defined on, the path of integration and things such as the singularity in the integrand at $x=1$. e.g. $\frac{1}{z}$ has a singularity (pole) at the origin $\int \frac{1}{z}dz$ produces different results for closed loops depending whether or not the loop encloses that singularity. $\endgroup$
    – PM.
    Commented Feb 13, 2020 at 22:11
  • $\begingroup$ Hi, thanks for the quick response. I just want to make sure I am understanding this correctly. The reason this approach can not be used for x>1 is that you will go from real to complex variables. I think I understand that. However, why can't one integrate, for example, from 2 to infinity with x beeing a complex number and then taking the real part to get the same answer as one would get with a hyperbolic substitution? Thanks in advance. $\endgroup$ Commented Feb 13, 2020 at 22:21

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