How do you find the integral of #f(x)=e^xcos2x# using integration by parts?

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Answer 1 · 2018-04-02T16:42:28

#inte^x*cos2x dx= 1/5*e^x(cos2x+2sin2x)+C#

#I= inte^x*cos2x dx# . Let #u=cos2x :. du= -2sin2x dx#

#dv= e^xdx :. v= e^x # . Integration by parts : #uv-intvdu#

#I= cos2x*e^x-int - e^x*2sin2x dx # or

#I= cos2x*e^x+2(int e^x*sin2x dx ) #

Let #u=sin2x :. du= 2cos2x dx, dv= e^xdx :. v= e^x #

#I= cos2x*e^x+2[sin2x*e^x-inte^x*2cos2x ] # or

#I= cos2x*e^x+2sin2x*e^x-4inte^x*cos2x #

#I= cos2x*e^x+2sin2x*e^x-4I#

By Transposing we get

#5I = cos2x*e^x+2sin2x*e^x# or

#5I = e^x(cos2x+2sin2x)#

#:. I = 1/5*e^x(cos2x+2sin2x)+C#

#inte^x*cos2x dx= 1/5*e^x(cos2x+2sin2x)+C# [Ans]