Consider the expression:
$$\int \cosh(x) \cdot \sinh(x) dx$$
Case 1. Solving the integral by substitution, I would set $s=\sinh(x)$. First we differentiate s:
$$s'_x=\frac{ds}{dx}\quad\Leftrightarrow\quad dx=\frac{1}{s'_x}ds=\frac{1}{\cosh(x)}ds$$
And then we can plug it into the integral:
$$\int \cosh(x) \cdot \sinh(x) dx=\int \cosh(x) \cdot s\; dx=\int \cosh(x) \cdot s\; \frac{1}{\cosh(x)}ds\\ =\int s\; ds=\frac12 s^2 =\frac12 \sinh(x)^2$$
Case 2. If I do the exact same thing, but instead set $s=\cosh(x)$ in the beginning, then I end up with:
$$\int \cosh(x) \cdot \sinh(x) dx =\frac12 \cosh(x)^2$$ Different result. I suspect that I'm doing some mistake with the borders of the integral. But can anyone point at it?
The "$+ \text{ constant}$" that's usually written after an indefinite integral isn't just for show. Your two solutions satisfy $\frac{1}{2} \cosh(x)^2 - \frac{1}{2} \sinh(x)^2 = \frac{1}{2}$, in other words they differ by a constant. This is perfectly fine: if $F$ is a primitive of $f$, then $F + c$ is a primitive of $f$ too, for any constant $c$. In this case the constant happens to be $\frac{1}{2}$.
Both $\frac12 \cosh(x)^2$ and $\frac12 \sinh(x)^2$ are primitives of $\cosh(x) \sinh(x)$: if you're not convinced, try differentiating both. (This probably should have been your first reflex if you weren't sure of your solution...)