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Question Number 95405 by i jagooll last updated on 25/May/20

$$\int {\mathrm{e}}^{\mathrm{x}}(\tan \mathrm{x}-\ln \left(\cos \mathrm{x}\right))\mathrm{dx}?$$

Commented by PRITHWISH SEN 2 last updated on 25/May/20

$$\mathrm{yes}$$

Commented by bobhans last updated on 25/May/20

$$\mathrm{no}.-{\mathrm{e}}^{\mathrm{x}}\ln \left(\cos \mathrm{x}\right)+\mathrm{c}$$$$\mathrm{or}{\mathrm{e}}^{\mathrm{x}}\ln \left(\sec \mathrm{x}\right)+\mathrm{c}$$

Commented by bobhans last updated on 25/May/20

$$\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{\mathrm{x}}\ln \left(\cos \mathrm{x}\right)\right)={\mathrm{e}}^{\mathrm{x}}\ln \left(\cos \mathrm{x}\right)+\frac{{\mathrm{e}}^{\mathrm{x}}(-\sin \mathrm{x})}{\cos \mathrm{x}}$$$$={\mathrm{e}}^{\mathrm{x}}\ln \left(\cos \mathrm{x}\right)-{\mathrm{e}}^{\mathrm{x}}\tan \left(\mathrm{x}\right)\ne \mathrm{with}\mathrm{equation}$$

Answered by bobhans last updated on 25/May/20

$$\int (\tan \mathrm{x}-\ln \left(\cos \mathrm{x}\right))\mathrm{d}\left({\mathrm{e}}^{\mathrm{x}}\right)=\mathrm{I}$$$$\mathrm{u}=\tan \mathrm{x}-\ln \left(\cos \mathrm{x}\right))$$$$\mathrm{du}=\sec {}^2\mathrm{x}+\tan \mathrm{x}\mathrm{dx}$$$$\mathrm{I}={\mathrm{e}}^{\mathrm{x}}(\tan \mathrm{x}-\ln \left(\cos \mathrm{x}\right))-\int {\mathrm{e}}^{\mathrm{x}}(\sec {}^2\mathrm{x}+\tan \mathrm{x})\mathrm{dx}$$$$\mathrm{I}={\mathrm{e}}^{\mathrm{x}}(\tan \mathrm{x}-\ln \left(\cos \mathrm{x}\right))-\int {\mathrm{e}}^{\mathrm{x}}\mathrm{d}\left(\tan \mathrm{x}\right)-\int {\mathrm{e}}^{\mathrm{x}}\tan \mathrm{x}\mathrm{dx}$$$$\mathrm{I}={\mathrm{e}}^{\mathrm{x}}(\tan \mathrm{x}-\ln \left(\cos \mathrm{x}\right))-[{\mathrm{e}}^{\mathrm{x}}\tan \mathrm{x}-\int {\mathrm{e}}^{\mathrm{x}}\tan \mathrm{x}\mathrm{dx}]-\int {\mathrm{e}}^{\mathrm{x}}\tan \mathrm{x}\mathrm{dx}$$$$\mathrm{I}={\mathrm{e}}^{\mathrm{x}}\tan \mathrm{x}-{\mathrm{e}}^{\mathrm{x}}\ln \left(\cos \mathrm{x}\right))-{\mathrm{e}}^{\mathrm{x}}\tan \mathrm{x}+\mathrm{c}$$$$\mathrm{I}={\mathrm{e}}^{\mathrm{x}}\ln \left(\sec \mathrm{x}\right)+\mathrm{c}$$

Commented by i jagooll last updated on 25/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}\mathrm{bob}$$

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