I'm trying to do the integral: $$\int{\frac{1}{\sqrt{e^{-2x}-1}}}dx$$ So I try two ways to do it, the first method I used is to multiply $e^x$ on both sides first. $$\int{\frac{1}{\sqrt{e^{-2x}-1}}}dx$$ $$=\int{\frac{e^x}{\sqrt{1-e^{2x}}}}dx$$ $$=\int{\frac{\sin{\theta}}{\sqrt{1-\sin^2{\theta}}}\times\frac{\cos{\theta}}{\sin{\theta}}}d\theta$$ $$=\theta+C$$ $$=\arcsin{(e^x)}+C$$ The second method is to do substitution directly. $$\int{\frac{1}{\sqrt{e^{-2x}-1}}}dx$$ $$=-\int{\frac{1}{u}\times\frac{u}{u^2+1}}du$$ $$=-\arctan{u}+C$$ $$=-\arctan{\sqrt{e^{-2x}-1}}+C$$ $\text{After input them into desmos, I found out that the distances betweens the graphs is constant}\frac{\pi}{2}$ $\text{So I want to ask that, can you prove that,} $ $$\arcsin{(e^x)}-(-\arctan{\sqrt{e^{-2x}-1}})\text{ is a constant?}$$
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3Hint: set $e^x= \sin \theta$ – Vasili Mar 21 '24 at 13:23
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5I understand what you are asking, but I still want to point out that, assuming your calculations are correct, you just proved that. – Carsten S Mar 22 '24 at 12:37
4 Answers
You can compute the derivatives:
$$\arcsin'y = \frac{1}{\sqrt{1-y^2}} ,$$ $$\arctan'y = \frac{1}{1+y^2} .$$ This shows that the derivative of $\arcsin{(e^x)}-(-\arctan{\sqrt{e^{-2x}-1}})$ is $$\frac{1}{\sqrt{1-e^{2x}}}e^x + \frac{1}{1+e^{-2x} -1}\frac 1 2 \frac{1}{\sqrt{e^{-2x}-1}}(-2e^{-2x}) = \frac{e^x}{\sqrt{1-e^{2x}}} - \frac{1}{\sqrt{e^{-2x}-1}} \\= \frac{e^x}{\sqrt{1-e^{2x}}} - \frac{e^x}{\sqrt{1 -e^{2x}}} = 0 .$$ Hence $\arcsin{(e^x)}-(-\arctan{\sqrt{e^{-2x}-1}})= C$.
With $x = 0$ we get $$C = \arcsin 1 + \arctan 0 = \frac{\pi}{2} + 0 = \frac{\pi}{2}.$$
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9I like this answer a lot. “How do I show this function is constant?” “Take its derivative and see if it is zero.” – Matthew Leingang Mar 21 '24 at 14:12
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1Why bother computing the derivatives? By the integration proof in the OP, you can just state that both functions have the same derivative. I don't think this answer adds anything. (sorry, I don't mean to be mean!) – preferred_anon Mar 22 '24 at 21:10
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@preferred_anon I disagree, the answer does add something, namely a general method that works in various scenarios, not only the specific scenario in the OP. – SampleTime Mar 23 '24 at 09:29
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@preferred_anon The question was "So I want to ask that, can you prove that $\arcsin{(e^x)}-(-\arctan{\sqrt{e^{-2x}-1}})$ is a constant?" You are right that this is implicit in the OP's computations, but I think he wanted to get an alternative confirmation. – Kritiker der Elche Mar 23 '24 at 10:32
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@KritikerderElche I agree with you - my claim is that this isn't really an alternative confirmation. OP proves that $F' = G'$ by computing the indefinite integral in two different ways. When I read this answer and look for an "a-ha" moment, I see two things cancel out, and say: Oh, of course, that's what I just integrated. Why do the algebra? – preferred_anon Mar 25 '24 at 09:15
Note that $$\sin^{-1}e^x = \tan^{-1} \frac{e^x}{\sqrt{1-e^{2x}}} = \frac{\pi}2- \tan^{-1} {\sqrt{e^{-2x}-1}} $$
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2IMO this should be the accepted answer, as it shows how the substitution $e^x=\sin\theta$ isn't necessary when proving the results equal. My answer just builds off of yours to show from there that the two result are equal. – CrSb0001 Mar 21 '24 at 13:50
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1Maybe just mention that we're using these two results: (1) $\sin^{-1} x = \tan^{-1} \frac{x}{\sqrt{1-x^2}}$; and (2) $\tan^{-1} x + \tan^{-1} \frac{1}{x} = \frac \pi 2$ (for $x>0$) – user103496 Mar 22 '24 at 01:58
You have already proved it, by doing those integrals! If you compute the same indefinite integral in multiple ways, then (provided your computations were correct) whatever answers you get must be equal up to a constant.
But in case you’re not quite confident of your original calculations, you can check them by differentiating the solutions and checking they give the same answer, which should be the body of your original integral. That is, if you can show $f'(x) = g'(x)$, that means $\frac{d}{dx}(f(x)-g(x)) = 0$, so the difference $f(x)-g(x)$ is constant. Paul Frost’s answer goes through the details of this for your functions.
(Here I’m assuming that the solutions $f(x)$, $g(x)$ are defined on a connected domain. If there are holes in their domain, typically coming from singularities in the integrand, like e.g. $\int \frac{1}{x}\mathrm{d}x$, then the different between the solutions is only locally constant — the solutions must still agree up to some constant within each connected component of their domain, but the constant can be different on different components.)
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From @Quanto's answer, we know that$$\arcsin(e^x)=\pi/2-\arctan\sqrt{e^{-2x}-1}$$Now, our expression becomes$$\pi/2\require{cancel}-\arctan\sqrt{e^{-2x}-1}-(-\arctan\sqrt{e^{-2x}-1})\\=\pi/2\cancel{-\arctan\sqrt{e^{-2x}-1}}\cancel{+\arctan\sqrt{e^{-2x}-1}}\\=\pi/2$$So the two results differ by just $\pi/2$, just like what you observed.
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