How to integrate $\int_{0}^{t}{\frac{\cos u}{\cosh^2 u}du}$? How to integrate $\int_{0}^{t}{\frac{\cos u}{\cosh^2 u}du}$?
I'm trying to use the integration by parts but it's impossible...
Is there an other way?

If you are just interested in the limit as $t\to +\infty$ (namely $\frac{\pi}{2\sinh\frac{\pi}{2}}$), there are better ways (namely the Fourier transform), but you may notice that
$$\begin{eqnarray*}\int_{0}^{t}\frac{\cos u}{\cosh^2 u}\,du&=&\left[\cos(u)\tanh(u)\right]_{0}^{t}+\int_{0}^{t}\tanh(u)\sin(u)\,du\\&=&\cos(t)\tanh(t)+\cos(t)-1+2\int_{0}^{t}\frac{\sin(u)}{1+e^{-2u}}\,du\\&=&-1+\frac{2\cos(t)}{1+e^{-2t}}+2\,\text{Im}\sum_{n\geq 0}(-1)^n\int_{0}^{t}e^{iu-2nu}\,du\\&=&-1+\frac{2\cos(t)}{1+e^{-2t}}+2\sum_{n\geq 0}(-1)^n\frac{1-e^{-2nt}\cos(t)-2ne^{-2nt}\sin(t)}{4n^2+1}\\&=&\frac{\pi}{2\sinh\frac{\pi}{2}}+\frac{2\cos(t)}{1+e^{-2t}}-2\sum_{n\geq 0}(-1)^n e^{-2nt}\frac{\cos(t)+2n\sin(t)}{4n^2+1}.\end{eqnarray*} $$
