Change of variables in trigonometric integral 
Question #1: How do I exactly deduce the expression for $d\varphi$?
Question #2: How do I exactly perform the change of variables? I think there are a lot of intermediate steps missing (the algebraic ones).
My thoughts on Q1: it holds that $\displaystyle \varphi = -i\log(-\tanh(\pi t) - i \mathrm{sech}(\pi t))$. Does this help?
 A: Solved the first Question:
$\displaystyle d\varphi = \frac{d}{dt}\left[-i \log\left( -\tanh (\pi t) - i \mathrm{sech}(\pi t) \right)\right]
      = -\pi\frac{\frac{i}{\cosh^2(\pi t)} + \frac{\sinh(\pi t)}{\cosh^2(\pi t)}}{\tanh (\pi t) + i \mathrm{sech}(\pi t)}
      = -\pi \frac{i + \sinh(\pi t)}{\sinh(\pi t)\cosh(\pi t) + i\cosh(\pi t)}
      = -\pi \frac{1}{\cosh(\pi t)}
      = -\pi \mathrm{sech}(\pi t)$
And the second one:
$\displaystyle \sin(\varphi)\cosh(\pi t) = \sin\left( -i\log\left( -\tanh (\pi t) - i \mathrm{sech}(\pi t) \right)\right) \cosh(\pi t)
      = \frac{1}{2i} \left[ -\tanh(\pi t) - i\mathrm{sech}(\pi t) + \frac{1}{\tanh(\pi t) + i\mathrm{sech}(\pi t) }\right]\cosh(\pi t)
      = \frac{1}{2i} \left[ \frac{\cosh(\pi t) - \tanh(\pi t) \sinh(\pi t) - 2i\tanh(\pi t) + \mathrm{sech}(\pi t)}{\tanh(\pi t) + i\mathrm{sech}(\pi t)}\right]
      = \frac{1}{2i} \left[ \frac{\cosh^2(\pi t) - \sinh^2(\pi t) - 2i \sinh(\pi t) + 1}{\sinh(\pi t) + i}\right]
      = \frac{1 - i\sinh(\pi t)}{i\sinh(\pi t) - 1}
      = -1$
