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Question Number 106809 by Study last updated on 07/Aug/20

$$\int e^{2x}{sine}^{x}dx=??$$

Answered by bemath last updated on 07/Aug/20

$$@\mathrm{bemath}@$$$$\int {\mathrm{e}}^{\mathrm{x}}\sin \left({\mathrm{e}}^{\mathrm{x}}\right)\mathrm{d}\left({\mathrm{e}}^{\mathrm{x}}\right)=\int \mathrm{w}\sin \mathrm{w}\mathrm{dw}$$$$=-\mathrm{wcos}\mathrm{w}-\sin \mathrm{w}+\mathrm{C}$$$$=-{\mathrm{e}}^{\mathrm{x}}\cos \left({\mathrm{e}}^{\mathrm{x}}\right)-\sin \left({\mathrm{e}}^{\mathrm{x}}\right)+\mathrm{C}$$

Commented by malwaan last updated on 07/Aug/20

$$-e^{x}cos\left(e^x\right)-sin\left(e^x\right)+C$$$$thankyousirbemath$$

Commented by bemath last updated on 07/Aug/20

$$\mathrm{oo}\mathrm{yes}.\mathrm{sorry}\mathrm{typo}$$

Answered by mathmax by abdo last updated on 07/Aug/20

$$\mathrm{I}=\int {\mathrm{e}}^{2\mathrm{x}}\sin \left({\mathrm{e}}^{\mathrm{x}}\right)\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{tbe}\mathrm{changement}{\mathrm{e}}^{\mathrm{x}}=\mathrm{t}\Rightarrow$$$$\mathrm{I}=\int {\mathrm{t}}^2\sin \left(\mathrm{t}\right)\frac{\mathrm{dt}}{\mathrm{t}}=\int \mathrm{tsint}\mathrm{dt}\mathrm{by}\mathrm{parts}\mathrm{u}=\mathrm{t}\mathrm{and}{\mathrm{v}}^{{}^{\prime }}=\mathrm{sint}\Rightarrow$$$$\mathrm{I}=-\mathrm{tcost}+\int \mathrm{cost}\mathrm{dt}=-\mathrm{tcost}+\mathrm{sint}+\mathrm{C}$$$$=-{\mathrm{e}}^{\mathrm{x}}\cos \left({\mathrm{e}}^{\mathrm{x}}\right)+\sin \left({\mathrm{e}}^{\mathrm{x}}\right)+\mathrm{C}$$

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