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Question Number 136935 by metamorfose last updated on 27/Mar/21

∫cos(x)e^(cos(x)+sin(x)) dx=...?

$$\int{cos}\left({x}\right){e}^{{cos}\left({x}\right)+{sin}\left({x}\right)} {dx}=...? \\ $$

Answered by liberty last updated on 27/Mar/21

∫ cos x.e^(cos x) .e^(sin x) dx = by parts { ((u=e^(cos x) du=−sin x.e^(cos x) dx)),((v=∫cos x.e^(sin x) dx=∫d(e^(sin x) )=e^(sin x) )) :} ∫ e^(cos x) d(e^(sin x) ) = e^(cos x+sin x) +∫sin xe^(cos x+sin x) dx (1−sin x)∫cos x.e^(cos x+sin x) dx = e^(cos x+sin x) ∫ cos x.e^(cos x+sin x) dx = ((e^(cos x+sin x) +C)/(1−sin x))

$$\int\:\mathrm{cos}\:\mathrm{x}.\mathrm{e}^{\mathrm{cos}\:\mathrm{x}} .\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} \:\mathrm{dx}\:= \\ $$$$\mathrm{by}\:\mathrm{parts} \begin{cases}{\mathrm{u}=\mathrm{e}^{\mathrm{cos}\:\mathrm{x}} \: \mathrm{du}=−\mathrm{sin}\:\mathrm{x}.\mathrm{e}^{\mathrm{cos}\:\mathrm{x}} \:\mathrm{dx}}\\{\mathrm{v}=\int\mathrm{cos}\:\mathrm{x}.\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} \:\mathrm{dx}=\int\mathrm{d}\left(\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} \right)=\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} }\end{cases} \\ $$$$\int\:\mathrm{e}^{\mathrm{cos}\:\mathrm{x}} \:\mathrm{d}\left(\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} \right)\:=\:\mathrm{e}^{\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}} +\int\mathrm{sin}\:\mathrm{xe}^{\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}} \:\mathrm{dx} \\ $$$$\left(\mathrm{1}−\mathrm{sin}\:\mathrm{x}\right)\int\mathrm{cos}\:\mathrm{x}.\mathrm{e}^{\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}} \:\mathrm{dx}\:=\:\mathrm{e}^{\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}} \\ $$$$\int\:\mathrm{cos}\:\mathrm{x}.\mathrm{e}^{\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}} \:\mathrm{dx}\:=\:\frac{\mathrm{e}^{\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}} \:+\mathrm{C}}{\mathrm{1}−\mathrm{sin}\:\mathrm{x}} \\ $$

Commented by metamorfose last updated on 28/Mar/21

Sir , i can′t understand the ligne 4

$${Sir}\:,\:{i}\:{can}'{t}\:{understand}\:{the}\:{ligne}\:\mathrm{4}\: \\ $$

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