I've been doing this question for a bit and I can't get my head around it.
I'm meant to evaluate $I = \int \frac1{\cosh(x)} dx$ by substitution, using $e^x$ as $u$.
So far I have,
$$\int \frac{2}{e^x+e^{-x}} dx$$
$u=e^x, du=e^xdx$
"$dx=du/e^x$" --Not sure if I can do this.
$$2\int \frac{\frac{1}{e^x}}{e^x+\frac{1}{e^x}} du\\ 2\int \frac{\frac{1}{u}}{u+\frac{1}{u}} du$$
And at this stage I get stuck. It looks kind of a function and its derivative I think? If I represent it as:
$$2\int u^{-1}(u+u^{-1})^{-1}$$
But not enough for me to be able to obviously recognise it and doctor it to use integration by substitution again. I think I'm overthinking the question and it's probably a lot simpler than this. Thanks heaps for any help.