$\int\frac{e^x}{\sqrt{1-e^{2x}}}\,dx$ with $e^x=\cos(u)$

Question

Consider the integral $$ I=\int\frac{e^x}{\sqrt{1-e^{2x}}}\,dx $$ When using the substitution $e^x=\sin(u)$, I get $I=\arcsin(e^x)+C$. However, when using $e^x=\cos(u)$, I get $du/dx=-e^x/\sin(u)$ and thus $$ \begin{align} \int\frac{e^x}{\sqrt{1-e^{2x}}}\,dx&=\int \frac{\cos (u)}{\sqrt{1-\cos^2(u)}}\frac{dx}{du}du\\ &=-\int \frac{\cos(u)}{\sin(u)}\frac{\sin(u)}{e^x}\,du\\ &=-\int 1 \,du=-u+C=-\arccos(e^x)+C. \end{align} $$ What am I missing?

Answer 1

The answers in question are both correct (the +C exists for a reason). Plot both graphs and you will see they are just offsets of each other (hence the constant)

The method to get to the answers, however, is questionable. $e^{x} = \sin(u)$ is impossible whenever $x>0$, as it will be outside sines range. Interestingly enough, it still arrives at the right answer.

Now, to actually solve this, the proper way would be to consider $u=e^{x}$, meaking $du = e^{x}dx$, and:

$\int\frac{e^x}{\sqrt{1-e^{2x}}}\,dx = \int{\frac{1}{\sqrt{1-u^{2}}}}du = \sin^{-1}(u) +C = \sin^{-1}(e^{x})+C$

EDIT The substitution is fine, as the denominator of the function guarantees that $x<0$, meaning the sub works. Sorry!