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Question Number 121046 by bounhome last updated on 05/Nov/20

$$1.\int e^{x}sinxdx=?$$$$$$$$2.\int e^{x}cosxdx=?$$

Answered by Ar Brandon last updated on 05/Nov/20

$$\mathrm{\Psi }=\int {\mathrm{e}}^{\mathrm{x}}{\mathrm{e}}^{\mathrm{xi}}\mathrm{dx}=\int {\mathrm{e}}^{\mathrm{x}+\mathrm{xi}}\mathrm{dx}=\frac{1}{1+\mathrm{i}}{\mathrm{e}}^{\mathrm{x}+\mathrm{xi}}+\mathrm{C}$$$$=\frac{{\mathrm{e}}^{\mathrm{x}}(\mathrm{cosx}+\mathrm{isinx})}{1+\mathrm{i}}+\mathrm{C}=\frac{{\mathrm{e}}^{\mathrm{x}}(\mathrm{cosx}+\mathrm{isinx})(1-\mathrm{i})}{2}+\mathrm{C}$$$$=\frac{{\mathrm{e}}^{\mathrm{x}}}{2}\{(\mathrm{cosx}+\mathrm{sinx})+\mathrm{i}(\mathrm{sinx}-\mathrm{cosx})\}+\mathrm{C}$$$$1.\int {\mathrm{e}}^{\mathrm{x}}\mathrm{sinxdx}=Im\int {\mathrm{e}}^{\mathrm{x}+\mathrm{ix}}\mathrm{dx}=\frac{{\mathrm{e}}^{\mathrm{x}}(\mathrm{sinx}-\mathrm{cosx})}{2}+k$$$$2.\int {\mathrm{e}}^{\mathrm{x}}\mathrm{cosxdx}=Re\int {\mathrm{e}}^{\mathrm{x}+\mathrm{ix}}\mathrm{dx}=\frac{{\mathrm{e}}^{\mathrm{x}}(\mathrm{cosx}+\mathrm{sinx})}{2}+\lambda$$

Answered by Ar Brandon last updated on 05/Nov/20

$$1.\mathcal{I}=\int {\mathrm{e}}^{\mathrm{x}}\mathrm{sinxdx}=\mathrm{sinx}\int {\mathrm{e}}^{\mathrm{x}}\mathrm{dx}-\int \{\frac{\mathrm{dsinx}}{\mathrm{dx}}\cdot \int {\mathrm{e}}^{\mathrm{x}}\}\mathrm{dx}$$$$={\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}-\int {\mathrm{e}}^{\mathrm{x}}\mathrm{cosxdx}={\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}-[{\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}+\int {\mathrm{e}}^{\mathrm{x}}\mathrm{sinxdx}]$$$$=\frac{{\mathrm{e}}^{\mathrm{x}}(\mathrm{sinx}-\mathrm{cosx})}{2}+k$$

Answered by ebi last updated on 05/Nov/20

$$1)$$$$I_1=\int e^{x}sinxdx$$$$let{u}_1=sinx\to {du}_1=cosxdx$$$${dv}_1=e^{x}dx\to v_1=e^x$$$$I_1=u_1{v}_1-\int v_{1}du$$$$I_1=e^{x}sinx-\int e^{x}cosxdx$$$$I_1=e^{x}sinx-I_2.....\left(1\right)$$$$$$$$I_2=\int e^{x}cosxdx$$$$let{u}_2=cosx\to {du}_2=-sinxdx$$$${dv}_2=e^{x}dx\to v_2=e^x$$$$I_2=u_2{v}_2-\int v_2{du}_2$$$$I_2=e^{x}cosx-\int e^x(-sinx)dx$$$$I_2=e^{x}cosx+\int e^{x}sinxdx.....\left(2\right)$$$$$$$$substitue\left(2\right)into\left(1\right)$$$$I_1=e^{x}sinx-(e^{x}cosx+\int e^{x}sinxdx)$$$$I_1=e^{x}sinx-e^{x}cosx-\int e^{x}sinxdx$$$$I_1=e^{x}sinx-e^{x}cosx-I_1$$$$2I_1=e^{x}sinx-e^{x}cosx$$$$I_1=\frac{e^x}{2}(sinx-cosx)$$$$\therefore \int e^{x}sinxdx=\frac{e^x}{2}(sinx-cosx)+c$$$$$$$$2)$$$$\int e^{x}cosxdx=\frac{e^x}{2}(cosx+sinx)+c$$

Answered by n0y0n last updated on 05/Nov/20

$$\mathrm{USE}\mathrm{DI}\mathrm{Method}$$

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