How to prove the two answers to an integral are equivalent

Question

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,} $ $$\arcsin{(e^x)}-(-\arctan{\sqrt{e^{-2x}-1}})\text{ is a constant?}$$

Answer 1

Note that $$\sin^{-1}e^x = \tan^{-1} \frac{e^x}{\sqrt{1-e^{2x}}} = \frac{\pi}2- \tan^{-1} {\sqrt{e^{-2x}-1}} $$

Answer 2

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.

Answer 3

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 $$C = \arcsin 1 + \arctan 0 = \frac{\pi}{2} + 0 = \frac{\pi}{2}.$$

Answer 4

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.)