Question Number 164367 by Zaynal last updated on 16/Jan/22
$$\int\:\boldsymbol{{e}}^{\boldsymbol{{tan}}\left(\boldsymbol{{x}}\right)} \:\boldsymbol{{dx}} \\ $$$$\left\{\boldsymbol{{Z}}.\boldsymbol{{A}}\right\} \\ $$
Answered by puissant last updated on 16/Jan/22
$$\Omega=\int{e}^{{tanx}} {dx}\:;\:{t}={tanx}\:\rightarrow\:{dx}=\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\Rightarrow\:\Omega=\int\frac{{e}^{{t}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\int\left\{\frac{{e}^{{t}} }{{t}−{i}}−\frac{{e}^{{t}} }{{t}+{i}}\right\}{dt} \\ $$$$=\:−\frac{{i}}{\mathrm{2}}\int\frac{{e}^{{i}} {e}^{{t}−{i}} }{{t}−{i}}{dt}\:+\:\frac{{i}}{\mathrm{2}}\int\frac{{e}^{−{i}} {e}^{{t}+{i}} }{{t}+{i}}{dt} \\ $$$$=\:−\frac{{ie}^{{i}} }{\mathrm{2}}{Ei}\left({t}−{i}\right)+\frac{{ie}^{−{i}} }{\mathrm{2}}{Ei}\left({t}+{i}\right)+{C} \\ $$$$=\:−\frac{{ie}^{{i}} }{\mathrm{2}}{Ei}\left({tanx}−{i}\right)+\frac{{ie}^{−{i}} }{\mathrm{2}}{Ei}\left({tanx}+{i}\right)+{C}. \\ $$$$\:\:\:\:\:……………\mathscr{L}{e}\:{puissant}…………. \\ $$