$$\int\frac{{e}^{{x}} }{{cosx}}{dx} \\$$
$$=\left[{e}^{{x}} ×\frac{\mathrm{1}}{{cosx}}\right]−\int{e}^{{x}} ×−\frac{{sinx}}{{cos}^{\mathrm{2}} {x}}{dx} \\$$Nous considérons U'=
$$\int{e}^{{x}} ×\frac{{sinx}}{{cos}^{\mathrm{2}} {x}}{dx}=\left[{tgx}×{e}^{{x}} {sinx}\right]−\int{tgx}\left({e}^{{x}} {sinx}+{e}^{{x}} {cosx}\right) \\$$$${I}=\left[\frac{{e}^{{x}} }{{cosx}}+{tgx}×{e}^{{x}} {sinx}\right]−\int{e}^{{x}} ×\frac{{sin}^{\mathrm{2}} {x}}{{cosx}}{dx}−\int{e}^{{x}} {sinxdx} \\$$$${l}\:{integrale}\:\int{e}^{{x}} {sinxdx}\:\:{etant}\:{simple} \\$$$${cherchons}\:\:\int{e}^{{x}} ×\frac{{sin}^{\mathrm{2}} {x}}{{cosx}}{dx} \\$$$$\int{e}^{{x}} ×\frac{\mathrm{1}−{cos}^{\mathrm{2}} {x}}{{cosx}}{dx} \\$$$$\int\frac{{e}^{{x}} }{{cosx}}−\int{e}^{{x}} {cosxdx} \\$$$${on}\:{obtient} \\$$$${I}=\left[{e}^{{x}} \left(\frac{\mathrm{1}}{{cosx}}+{tgx}×{sinx}\right)\right]−\int{e}^{{x}} {sinxdx}−\left(\int\frac{{e}^{{x}} }{{cosx}}−\int{e}^{{x}} {cosx}\right) \\$$$$\mathrm{2}{I}=\left[{e}^{{x}} \left(\frac{\mathrm{1}}{{cosx}}+\frac{{sin}^{\mathrm{2}} {x}}{{cosx}}\right)\right]+\int{e}^{{x}} {cosxdx}−\int{e}^{{x}} {sinxdx} \\$$$$\int{e}^{{x}} {cosxdx}=\left[{e}^{{x}} {cosx}\right]+\int{e}^{{x}} {sinxdx} \\$$$$\mathrm{2}{I}=\left[{e}^{{x}} \left(\frac{\mathrm{1}+{sin}^{\mathrm{2}} {x}}{{cosx}}\right)+{cosx}\right] \\$$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\:\left[{e}^{{x}} \left(\frac{\mathrm{1}+{sin}^{\mathrm{2}} {x}}{{cosx}}\right)+{cosx}\right]+{cte} \\$$