Let $I= \int\operatorname{sech}x dx$ .
On the one hand when we apply the substitution $u = e^x$ we get equation 1: $I \stackrel{(1)}{=} 2\arctan(e^x)+c$. We could also substitute to $u = \tanh(x)$ and obtain 2: $I \stackrel{(2)}{=} \arcsin(\operatorname{sech}(x))+c$. Or we could substitute $u = \sinh(x)$ which makes 3: $I\stackrel{(3)}{=} \arctan(\sinh(x))+c$. How can we show the equivalence, at least by a constant for each of these results? I'm bewildered by equations 2 and 3 because $\arctan(t)= \arcsin(\frac{t}{\sqrt{1+t^2}})$, which implies $I=\arcsin(\tanh(x))+c$.
Either I've gone wrong somewhere or every one of these answers is right.
Write $I_1=2\arctan(e^x), I_2=\arcsin(\mathrm{sech}(x)), I_3=\arctan(\sinh(x))$. Then $$I_1 = I_3+\frac{\pi}{2}$$ and (with Maple's conventions for principal value of $\arcsin$) $$ I_2 = I_1\quad\text{for } x \le 0 \\ I_2 = \pi - I_1 \quad\text{for } x > 0 $$ So I conclude $I_1$ and $I_3$ are correct, and $I_2$ is correct up to sign.