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I thought that $\theta$ must be chosen such that $\cosh(\theta)$ has a range that is equal to the domain of $\frac{1}{(x^2-1)^{3/2}}$. But this can't be done since the domain includes negative numbers, and $\cosh(\theta)$ is always positive. If this substitution is made, the answer seems to be valid for all values of x, why does it work?

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  • $\begingroup$ It's better to use $x=\sec t,\sqrt{x^2-1}=|\tan t|$ $\endgroup$ – lab bhattacharjee Nov 20 '18 at 14:15
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So there are singularities at $\pm 1$, you have to avoid those. So let's consider finding an antiderivative valid on $(-\infty,-1)$.

In fact $x=\cosh(\theta)$ is not valid there. The substitution really being used there is $x=-\cosh(\theta)$, so $dx=-\sinh(\theta) d \theta$, so the integral becomes

$$\int \frac{1}{(\cosh^2(\theta)-1)^{3/2}} (-\sinh(\theta)) d \theta.$$

The catch comes when you rewrite $(\cosh^2(\theta)-1)^{3/2}$ as just $\sinh^3(\theta)$. This is not strictly correct, in fact it is $|\sinh^3(\theta)|$ in general. On $(-\infty,-1)$, this absolute value reduces to $-\sinh^3(\theta)$. The formula collapses back to the formula you would get for an antiderivative on $(1,\infty)$ because this minus sign cancels out with the minus sign that came with $dx$.

This is relatively general, but I'm not sure whether pinning down exactly what happens in general will be helpful to you at this point.

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  • $\begingroup$ Thank you. This response had a lot of detail and answered my question. $\endgroup$ – DinW Nov 20 '18 at 21:31
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If you look for a primitive in $(1,+\infty)$,

put $$x=\cosh(\theta)$$

with $$dx=\sinh(\theta) d\theta$$

but if you want a primitive in $(-\infty,-1)$, you should put $$x=-\cosh(\theta)$$

with $$dx=-\sinh(\theta)d\theta$$

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Part of it's because we're presumably only integrating over values of $x$ for which the integrand is real, i.e. $|x|\ge 1$. Any work with complex numbers that lets you pose the question otherwise in the first place also lets you get a complex $\operatorname{arcosh}x$ with $|x|<1$.

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$$I=\int\left(x^2-1\right)^{-3/2}dx$$ we want this term $x^2-1$ to be simplified to a single term, so think of trig identities.: $$\cos^2x+\sin^2x\equiv1,\,\cosh^2x-\sinh^2x\equiv1$$ are the ones you need to know. since it is of the form $x^2-b^2$ rather than $b^2-x^2$ or $x^2+b^2$ we know to use hyperbolic functions. Manipulating the right rule, we obtain: $$\sinh^2x=\cosh^2x-1$$ now we can use this. let: $x=\cosh(t)$ and we get $dx=\sinh(t)$ and rewrite the integral as: $$I=\int\left(\cosh^2(t)-1\right)^{-3/2}.\sinh(t)dt=\int(sinh(t))^{-3}.\sinh(t)dt=\int\text{csch}^2(t)dt=-\coth(t)+C$$

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  • $\begingroup$ Hi, thanks for the response. I asking more about the choice for the substitution than how to solve the integral itself. Since, the range of cosh(x) doesn't equal the domain of the function that is being integrated. $\endgroup$ – DinW Nov 21 '18 at 1:22
  • $\begingroup$ It is a known option in examples like these where you have a fractional power of a second order polynomial $\endgroup$ – Henry Lee Nov 21 '18 at 1:23

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