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Question Number 164367 by Zaynal last updated on 16/Jan/22

∫ e^(tan(x)) dx {Z.A}

$$\int\:\boldsymbol{{e}}^{\boldsymbol{{tan}}\left(\boldsymbol{{x}}\right)} \:\boldsymbol{{dx}} \\ $$$$\left\{\boldsymbol{{Z}}.\boldsymbol{{A}}\right\} \\ $$

Answered by puissant last updated on 16/Jan/22

Ω=∫e^(tanx) dx ; t=tanx → dx=(1/(1+t^2 ))dt ⇒ Ω=∫(e^t /(1+t^2 ))dt = (1/(2i))∫{(e^t /(t−i))−(e^t /(t+i))}dt = −(i/2)∫((e^i e^(t−i) )/(t−i))dt + (i/2)∫((e^(−i) e^(t+i) )/(t+i))dt = −((ie^i )/2)Ei(t−i)+((ie^(−i) )/2)Ei(t+i)+C = −((ie^i )/2)Ei(tanx−i)+((ie^(−i) )/2)Ei(tanx+i)+C. ...............Le puissant.............

$$\Omega=\int{e}^{{tanx}} {dx}\:;\:{t}={tanx}\:\rightarrow\:{dx}=\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\Rightarrow\:\Omega=\int\frac{{e}^{{t}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\int\left\{\frac{{e}^{{t}} }{{t}−{i}}−\frac{{e}^{{t}} }{{t}+{i}}\right\}{dt} \\ $$$$=\:−\frac{{i}}{\mathrm{2}}\int\frac{{e}^{{i}} {e}^{{t}−{i}} }{{t}−{i}}{dt}\:+\:\frac{{i}}{\mathrm{2}}\int\frac{{e}^{−{i}} {e}^{{t}+{i}} }{{t}+{i}}{dt} \\ $$$$=\:−\frac{{ie}^{{i}} }{\mathrm{2}}{Ei}\left({t}−{i}\right)+\frac{{ie}^{−{i}} }{\mathrm{2}}{Ei}\left({t}+{i}\right)+{C} \\ $$$$=\:−\frac{{ie}^{{i}} }{\mathrm{2}}{Ei}\left({tanx}−{i}\right)+\frac{{ie}^{−{i}} }{\mathrm{2}}{Ei}\left({tanx}+{i}\right)+{C}. \\ $$$$\:\:\:\:\:...............\mathscr{L}{e}\:{puissant}............. \\ $$

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