How do you integrate #int e^x sin x dx # using integration by parts?

1 Answer
Lovecraft
Jan 7, 2016

#inte^xsin(x)dx = (e^xsin(x) - e^xcos(x))/2 + c#

Explanation:

Say #dv = e^x# so #v = e^x#, #u = sin(x)# so #du = cos(x)#

#inte^xsin(x)dx = e^xsin(x) - inte^(x)cos(x)dx#

Say #dv = e^x# so #v = e^x#, #u = cos(x)# so #du = -sin(x)#

#inte^xsin(x)dx = e^xsin(x) - (e^xcos(x) +inte^xsin(x)dx)#
#inte^xsin(x)dx = e^xsin(x) - e^xcos(x) -inte^xsin(x)dx#
#2inte^xsin(x)dx = e^xsin(x) - e^xcos(x)#
#inte^xsin(x)dx = (e^xsin(x) - e^xcos(x))/2 + c#

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