# Doing a standard integral with complex numbers instead of using a trigonometric substitution

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?

• $$\int\frac{1}{\sqrt{x^2-1}}dx=\text{arcosh}(x)+C=\pm i\arccos(x)+C$$ $$=\pm i\left(\frac\pi2-\arcsin(x)\right)+C=\mp i\arcsin(x)+\left(C\pm i\frac\pi2\right)$$ $$=\pm\frac1i\arcsin(x)+D$$ Commented Feb 13, 2020 at 22:55
• @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! Commented Feb 13, 2020 at 23:45
• $$\cosh(\pm i\theta)=\sum_{k=0}^\infty\frac{(\pm i\theta)^{2k}}{(2k)!}=\sum_{k=0}^\infty(-1)^k\frac{\theta^{2k}}{(2k)!}=\cos(\theta)$$ Commented Feb 13, 2020 at 23:48

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 $$-1 \le x \le 1 \Rightarrow x^2 \le 1$$
• 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. Commented Feb 13, 2020 at 21:33
• 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.