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Question Number 169253 by Mastermind last updated on 27/Apr/22

∫((−sinx)/e^x )dx    Mastermind

$$\int\frac{−{sinx}}{{e}^{{x}} }{dx} \\ $$$$ \\ $$$${Mastermind} \\ $$

Answered by thfchristopher last updated on 27/Apr/22

=-∫e^(-x) sin xdx  =∫sin xd(e^(-x) )  =e^(-x) sin x−∫e^(-x) d(sin x)  =e^(-x) sin x−∫e^(-x) cos xdx  =e^(-x) sin x+∫cos xd(e^(-x) )  =e^(-x) sin x+e^(-x) cos x−∫e^(-x) d(cos x)  =e^(-x) (sin x+cos x)+∫e^(-x) sin xdx  ∴-2∫e^(-x) sin xdx=e^(-x) (sin x+cos x)  -∫e^(-x) sin xdx=(1/2)e^(-x) (sin x+cos x)+C

$$=-\int{e}^{-{x}} \mathrm{sin}\:{xdx} \\ $$$$=\int\mathrm{sin}\:{xd}\left({e}^{-{x}} \right) \\ $$$$={e}^{-{x}} \mathrm{sin}\:{x}−\int{e}^{-{x}} {d}\left(\mathrm{sin}\:{x}\right) \\ $$$$={e}^{-{x}} \mathrm{sin}\:{x}−\int{e}^{-{x}} \mathrm{cos}\:{xdx} \\ $$$$={e}^{-{x}} \mathrm{sin}\:{x}+\int\mathrm{cos}\:{xd}\left({e}^{-{x}} \right) \\ $$$$={e}^{-{x}} \mathrm{sin}\:{x}+{e}^{-{x}} \mathrm{cos}\:{x}−\int{e}^{-{x}} {d}\left(\mathrm{cos}\:{x}\right) \\ $$$$={e}^{-{x}} \left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)+\int{e}^{-{x}} \mathrm{sin}\:{xdx} \\ $$$$\therefore-\mathrm{2}\int{e}^{-{x}} \mathrm{sin}\:{xdx}={e}^{-{x}} \left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right) \\ $$$$-\int{e}^{-{x}} \mathrm{sin}\:{xdx}=\frac{\mathrm{1}}{\mathrm{2}}{e}^{-{x}} \left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)+{C} \\ $$

Commented by Mastermind last updated on 28/Apr/22

Thanks sir, i really do appreciate

$${Thanks}\:{sir},\:{i}\:{really}\:{do}\:{appreciate} \\ $$

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