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Question Number 131706 by bounhome last updated on 07/Feb/21

$$\int e^{3x}cosxdx=?helpmeplease$$

Answered by mathmax by abdo last updated on 07/Feb/21

$$\int {\mathrm{e}}^{3\mathrm{x}}\mathrm{cosx}\mathrm{dx}=\mathrm{Re}(\int {\mathrm{e}}^{3\mathrm{x}}{\mathrm{e}}^{\mathrm{ix}}\mathrm{dx})=\mathrm{Re}(\int {\mathrm{e}}^{(3+\mathrm{i})\mathrm{x}}\mathrm{dx})\mathrm{we}\mathrm{have}$$$$\int {\mathrm{e}}^{(3+\mathrm{i})\mathrm{x}}\mathrm{ex}=\frac{1}{3+\mathrm{i}}{\mathrm{e}}^{(3+\mathrm{i})\mathrm{x}}+\mathrm{c}=\frac{3-\mathrm{i}}{10}{\mathrm{e}}^{3\mathrm{x}}(\mathrm{cosx}+\mathrm{isinx})$$$$=\frac{{\mathrm{e}}^{3\mathrm{x}}}{10}(3\mathrm{cosx}+3\mathrm{isinx}+\mathrm{icosx}+\mathrm{sinx})\Rightarrow$$$$\int {\mathrm{e}}^{3\mathrm{x}}\mathrm{cosx}\mathrm{ex}=\frac{{\mathrm{e}}^{3\mathrm{x}}}{10}(3\mathrm{cosx}+\mathrm{sinx})+\mathrm{C}$$

Answered by Dwaipayan Shikari last updated on 07/Feb/21

$$\frac{1}{2}\int e^{3x+ix}+e^{3x-ix}dx$$$$=\frac{1}{2}.\frac{e^{3x+ix}}{3+i}+\frac{1}{2}.\frac{e^{3x-ix}}{3-i}$$$$=\frac{e^{3x}}{2}(\frac{e^{ix}(3-i)}{10}+\frac{e^{-ix}(3+i)}{10})=\frac{e^{3x}}{10}(\frac{3}{2}(e^{ix}+e^{-ix}-\frac{1}{2}i(e^{ix}-e^{-ix}))$$$$=\frac{e^{3x}}{10}(3cosx+sinx)+C$$

Answered by physicstutes last updated on 07/Feb/21

$$\mathrm{let}\mathcal{I}=\int e^{3x}\cos xdx$$$$\mathrm{let}u=e^{3x}\mathrm{and}dv=\cos xdx$$$$\Rightarrow du=3e^{3x}dx\mathrm{and}v=\sin x$$$$\mathcal{I}=e^{3x}\sin x-3\int e^{3x}\sin xdx.........\left(1\right)$$$$\mathrm{consider}\mathcal{J}=\int e^{3x}\sin xdx$$$$\mathrm{again}\mathrm{let}u=e^{3x}\mathrm{and}dv=\sin xdx$$$$\Rightarrow du=3e^{3x}\mathrm{and}v=-\cos x$$$$\mathcal{J}=-e^{3x}\cos x+3\int e^{3x}\cos xdx..........\left(2\right)$$$$\left(2\right)\mathrm{in}\left(1\right)\Rightarrow \mathcal{I}=e^{3x}\sin x-3(-e^{3x}\cos x+3\int e^{3x}\cos xdx)$$$$\int e^{3x}\cos xdx=e^{3x}\sin x+3e^{3x}\cos x-9\int e^{3x}\cos xdx$$$$\Rightarrow \int e^{3x}\cos xdx=\frac{e^{3x}}{10}(\sin x+3\cos x)+A,\mathrm{where}A\in \mathbb{R}$$

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