How do you integrate $\int cos x \cdot e^{- x}$ $int cosx*e^-x$ by integration by parts method?

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Answer 1 · 2016-08-03T14:45:55

$\frac{1}{2} e^{- x} (sin x - cos x) + C$ $1/2 e^-x ( sin x - cos x ) + C$

$I = int cosx*e^-x \ dx$

$= int d/dx(sinx )e^-x \ dx$

by IBP
$= e^-x sin x- int sin x d/dx( e^-x) \ dx$

$= e^-x sin x + int sin x * e^-x \ dx$

$= e^-x sin x + int d/dx ( - cos x) e^-x \ dx$

$= e^-x sin x + (( - cos x) e^-x - int ( - cos x) d/dx ( e^-x) \ dx )$

$= e^-x sin x - e^-x cos x - int cos x * e^-x \ dx$

$= e^-x (sin x - cos x) - I + C$ adding an integation constant

$implies I =1/2 e^-x ( sin x - cos x ) + C$

How do you integrate int cosx*e^-x∫cosx⋅e−xint cosx*e^-x by integration by parts method?