How do you integrate #int cosx*e^-x# by integration by parts method?

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Answer 1 · 2016-08-03T14:45:55

#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 #