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

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Answer 1 · 2018-05-25T13:43:23

#int sinx e^(-x) dx = -(e^(-x)(sinx +cosx ))/2+ C#

Integrate by parts:

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

#int sinx e^(-x) dx = -e^(-x)sinx + int e^(-x) d/dx(sinx ) dx#

#int sinx e^(-x) dx = -e^(-x)sinx + int e^(-x) cosx dx#

and then again:

#int sinx e^(-x) dx = -e^(-x)sinx + int cosx d/dx (-e^(-x))dx#

#int sinx e^(-x) dx = -e^(-x)sinx -e^(-x)cosx + int e^(-x) d/dx(cosx )dx#

#int sinx e^(-x) dx = -e^(-x)sinx -e^(-x)cosx - int sinx e^(-x) dx#

The same integral appears on both sides and we can solve fro it:

#2int sinx e^(-x) dx = -e^(-x)sinx -e^(-x)cosx + C#

#int sinx e^(-x) dx = -(e^(-x)(sinx +cosx ))/2+ C#