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Question Number 33644 by mondodotto@gmail.com last updated on 21/Apr/18

 ∫(sec^2 x)e^(tanx) dx

$$\:\int\left(\boldsymbol{\mathrm{sec}}^{\mathrm{2}} \boldsymbol{{x}}\right)\boldsymbol{{e}}^{\boldsymbol{\mathrm{tan}{x}}} \boldsymbol{{dx}} \\ $$

Answered by Joel578 last updated on 21/Apr/18

I = ∫ (sec^2  x) e^(tan x)  dx  u = tan x  →  du = sec^2  x dx    I = ∫ e^u  du     = e^u  + C   = e^(tan x)  + C

$${I}\:=\:\int\:\left(\mathrm{sec}^{\mathrm{2}} \:{x}\right)\:{e}^{\mathrm{tan}\:{x}} \:{dx} \\ $$$${u}\:=\:\mathrm{tan}\:{x}\:\:\rightarrow\:\:{du}\:=\:\mathrm{sec}^{\mathrm{2}} \:{x}\:{dx} \\ $$$$ \\ $$$${I}\:=\:\int\:{e}^{{u}} \:{du} \\ $$$$\:\:\:=\:{e}^{{u}} \:+\:{C}\:\:\:=\:{e}^{\mathrm{tan}\:{x}} \:+\:{C} \\ $$

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