Question Number 77887 by aliesam last updated on 11/Jan/20
$$\int\frac{{e}^{{sinh}\left({x}\right)} }{{cosh}\left({x}\right)}\:{dx} \\ $$
Answered by MJS last updated on 12/Jan/20
![∫(e^(sinh x) /(cosh x))dx= [t=sinh x → dx=(1/(cosh x))] =∫(e^t /(t^2 +1))dt=∫(e^t /((t−i)(t+i)))dt= =(i/2)∫(e^t /(t+i))dt−(i/2)∫(e^t /(t−i))dt= [substitute u, v=t±i → du, dv=dt] =((ie^(−i) )/2)∫(e^u /u)du−((ie^i )/2)∫(e^v /v)dv= =((ie^(−i) )/2)Ei u −((ie^i )/2)Ei v = =((ie^(−i) )/2)Ei (t+i) −((ie^i )/2)Ei (t−i) = =((ie^(−i) )/2)Ei (sinh x +i) −((ie^i )/2)Ei (sinh x −i) +C](https://www.tinkutara.com/question/Q77986.png)
$$\int\frac{\mathrm{e}^{\mathrm{sinh}\:{x}} }{\mathrm{cosh}\:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{sinh}\:{x}\:\rightarrow\:{dx}=\frac{\mathrm{1}}{\mathrm{cosh}\:{x}}\right] \\ $$$$=\int\frac{\mathrm{e}^{{t}} }{{t}^{\mathrm{2}} +\mathrm{1}}{dt}=\int\frac{\mathrm{e}^{{t}} }{\left({t}−\mathrm{i}\right)\left({t}+\mathrm{i}\right)}{dt}= \\ $$$$=\frac{\mathrm{i}}{\mathrm{2}}\int\frac{\mathrm{e}^{{t}} }{{t}+\mathrm{i}}{dt}−\frac{\mathrm{i}}{\mathrm{2}}\int\frac{\mathrm{e}^{{t}} }{{t}−\mathrm{i}}{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{substitute}\:{u},\:{v}={t}\pm\mathrm{i}\:\rightarrow\:{du},\:{dv}={dt}\right] \\ $$$$=\frac{\mathrm{ie}^{−\mathrm{i}} }{\mathrm{2}}\int\frac{\mathrm{e}^{{u}} }{{u}}{du}−\frac{\mathrm{ie}^{\mathrm{i}} }{\mathrm{2}}\int\frac{\mathrm{e}^{{v}} }{{v}}{dv}= \\ $$$$=\frac{\mathrm{ie}^{−\mathrm{i}} }{\mathrm{2}}\mathrm{Ei}\:{u}\:−\frac{\mathrm{ie}^{\mathrm{i}} }{\mathrm{2}}\mathrm{Ei}\:{v}\:= \\ $$$$=\frac{\mathrm{ie}^{−\mathrm{i}} }{\mathrm{2}}\mathrm{Ei}\:\left({t}+\mathrm{i}\right)\:−\frac{\mathrm{ie}^{\mathrm{i}} }{\mathrm{2}}\mathrm{Ei}\:\left({t}−\mathrm{i}\right)\:= \\ $$$$=\frac{\mathrm{ie}^{−\mathrm{i}} }{\mathrm{2}}\mathrm{Ei}\:\left(\mathrm{sinh}\:{x}\:+\mathrm{i}\right)\:−\frac{\mathrm{ie}^{\mathrm{i}} }{\mathrm{2}}\mathrm{Ei}\:\left(\mathrm{sinh}\:{x}\:−\mathrm{i}\right)\:+{C} \\ $$