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Question Number 111010 by mohammad17 last updated on 01/Sep/20

$$\int e^{x}tanxdx$$

Answered by Rio Michael last updated on 02/Sep/20

$$\mathrm{Let}\mathrm{me}\mathrm{try}\mathrm{this},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{something}\mathrm{i}\mathrm{learnt}$$$$\mathrm{on}\mathrm{my}\mathrm{own}.$$$$\mathrm{Let}I=\int e^x\tan xdx.$$$$\mathrm{we}\mathrm{define}\tan x=-i\frac{1-e^{-2ix}}{1+e^{-2ix}}$$$$=-i-2i\underset{r=1}{\sum ^{\mathrm{\infty }}}{(-1)}^r{e}^{-2irx}$$$$\mathrm{so},\int e^x\tan xdx=-{ie}^x-2i\underset{r=1}{\sum ^{\mathrm{\infty }}}{(-1)}^r\int e^{(1-2ir)x}dx$$$$=-{ie}^x-2i\underset{r=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^r}{1-2ir}{e}^{(1-2ir)x}+k\left(\mathrm{a}\mathrm{constant}\right)$$$$\mathrm{am}\mathrm{sorry}\mathrm{but}\mathrm{this}\mathrm{solution}\mathrm{involves}\mathrm{the}L\mathrm{erch}P\mathrm{hi}\mathrm{function}\mathrm{which}$$$$\mathrm{i}\mathrm{have}\mathrm{no}\mathrm{understanding}\mathrm{of}\mathrm{it}\mathrm{till}\mathrm{today}.$$

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