Question Number 142389 by alcohol last updated on 31/May/21
$$\int\frac{{e}^{{x}} }{{cosx}}{dx} \\ $$
Answered by ArielVyny last updated on 31/May/21
![=[e^x ×(1/(cosx))]−∫e^x ×−((sinx)/(cos^2 x))dx Nous considérons U'= ∫e^x ×((sinx)/(cos^2 x))dx=[tgx×e^x sinx]−∫tgx(e^x sinx+e^x cosx) I=[(e^x /(cosx))+tgx×e^x sinx]−∫e^x ×((sin^2 x)/(cosx))dx−∫e^x sinxdx l integrale ∫e^x sinxdx etant simple cherchons ∫e^x ×((sin^2 x)/(cosx))dx ∫e^x ×((1−cos^2 x)/(cosx))dx ∫(e^x /(cosx))−∫e^x cosxdx on obtient I=[e^x ((1/(cosx))+tgx×sinx)]−∫e^x sinxdx−(∫(e^x /(cosx))−∫e^x cosx) 2I=[e^x ((1/(cosx))+((sin^2 x)/(cosx)))]+∫e^x cosxdx−∫e^x sinxdx ∫e^x cosxdx=[e^x cosx]+∫e^x sinxdx 2I=[e^x (((1+sin^2 x)/(cosx)))+cosx] I=(1/2) [e^x (((1+sin^2 x)/(cosx)))+cosx]+cte](https://www.tinkutara.com/question/Q142432.png)
$$=\left[{e}^{{x}} ×\frac{\mathrm{1}}{{cosx}}\right]−\int{e}^{{x}} ×−\frac{{sinx}}{{cos}^{\mathrm{2}} {x}}{dx} \\ $$Nous considérons U'=
$$\int{e}^{{x}} ×\frac{{sinx}}{{cos}^{\mathrm{2}} {x}}{dx}=\left[{tgx}×{e}^{{x}} {sinx}\right]−\int{tgx}\left({e}^{{x}} {sinx}+{e}^{{x}} {cosx}\right) \\ $$$${I}=\left[\frac{{e}^{{x}} }{{cosx}}+{tgx}×{e}^{{x}} {sinx}\right]−\int{e}^{{x}} ×\frac{{sin}^{\mathrm{2}} {x}}{{cosx}}{dx}−\int{e}^{{x}} {sinxdx} \\ $$$${l}\:{integrale}\:\int{e}^{{x}} {sinxdx}\:\:{etant}\:{simple} \\ $$$${cherchons}\:\:\int{e}^{{x}} ×\frac{{sin}^{\mathrm{2}} {x}}{{cosx}}{dx} \\ $$$$\int{e}^{{x}} ×\frac{\mathrm{1}−{cos}^{\mathrm{2}} {x}}{{cosx}}{dx} \\ $$$$\int\frac{{e}^{{x}} }{{cosx}}−\int{e}^{{x}} {cosxdx} \\ $$$${on}\:{obtient} \\ $$$${I}=\left[{e}^{{x}} \left(\frac{\mathrm{1}}{{cosx}}+{tgx}×{sinx}\right)\right]−\int{e}^{{x}} {sinxdx}−\left(\int\frac{{e}^{{x}} }{{cosx}}−\int{e}^{{x}} {cosx}\right) \\ $$$$\mathrm{2}{I}=\left[{e}^{{x}} \left(\frac{\mathrm{1}}{{cosx}}+\frac{{sin}^{\mathrm{2}} {x}}{{cosx}}\right)\right]+\int{e}^{{x}} {cosxdx}−\int{e}^{{x}} {sinxdx} \\ $$$$\int{e}^{{x}} {cosxdx}=\left[{e}^{{x}} {cosx}\right]+\int{e}^{{x}} {sinxdx} \\ $$$$\mathrm{2}{I}=\left[{e}^{{x}} \left(\frac{\mathrm{1}+{sin}^{\mathrm{2}} {x}}{{cosx}}\right)+{cosx}\right] \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\:\left[{e}^{{x}} \left(\frac{\mathrm{1}+{sin}^{\mathrm{2}} {x}}{{cosx}}\right)+{cosx}\right]+{cte} \\ $$
$$\int{e}^{{x}} ×\frac{{sinx}}{{cos}^{\mathrm{2}} {x}}{dx}=\left[{tgx}×{e}^{{x}} {sinx}\right]−\int{tgx}\left({e}^{{x}} {sinx}+{e}^{{x}} {cosx}\right) \\ $$$${I}=\left[\frac{{e}^{{x}} }{{cosx}}+{tgx}×{e}^{{x}} {sinx}\right]−\int{e}^{{x}} ×\frac{{sin}^{\mathrm{2}} {x}}{{cosx}}{dx}−\int{e}^{{x}} {sinxdx} \\ $$$${l}\:{integrale}\:\int{e}^{{x}} {sinxdx}\:\:{etant}\:{simple} \\ $$$${cherchons}\:\:\int{e}^{{x}} ×\frac{{sin}^{\mathrm{2}} {x}}{{cosx}}{dx} \\ $$$$\int{e}^{{x}} ×\frac{\mathrm{1}−{cos}^{\mathrm{2}} {x}}{{cosx}}{dx} \\ $$$$\int\frac{{e}^{{x}} }{{cosx}}−\int{e}^{{x}} {cosxdx} \\ $$$${on}\:{obtient} \\ $$$${I}=\left[{e}^{{x}} \left(\frac{\mathrm{1}}{{cosx}}+{tgx}×{sinx}\right)\right]−\int{e}^{{x}} {sinxdx}−\left(\int\frac{{e}^{{x}} }{{cosx}}−\int{e}^{{x}} {cosx}\right) \\ $$$$\mathrm{2}{I}=\left[{e}^{{x}} \left(\frac{\mathrm{1}}{{cosx}}+\frac{{sin}^{\mathrm{2}} {x}}{{cosx}}\right)\right]+\int{e}^{{x}} {cosxdx}−\int{e}^{{x}} {sinxdx} \\ $$$$\int{e}^{{x}} {cosxdx}=\left[{e}^{{x}} {cosx}\right]+\int{e}^{{x}} {sinxdx} \\ $$$$\mathrm{2}{I}=\left[{e}^{{x}} \left(\frac{\mathrm{1}+{sin}^{\mathrm{2}} {x}}{{cosx}}\right)+{cosx}\right] \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\:\left[{e}^{{x}} \left(\frac{\mathrm{1}+{sin}^{\mathrm{2}} {x}}{{cosx}}\right)+{cosx}\right]+{cte} \\ $$