How do you find the indefinite integral of ∫sin 4xe^sin2x dx?

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Answer 1 · 2018-04-13T09:03:50

#I=e^(sin(2x))(sin(2x)-1)+C#

We want to solve

#I=intsin(4x)e^(sin(2x))dx#

Using the trig identity #color(blue)(sin(2a)=2cos(a)sin(a)#

#I=2intcos(2x)sin(2x)e^(sin(2x))dx#

Make a substitution #u=sin(2x)=>du=2cos(2x)dx#

#I=intue^(u)du#

#I=ue^u-inte^udu#

#color(white)(I)=ue^u-e^u+C#

#color(white)(I)=e^u(u-1)+C#

Substitute back #u=sin(2x)#

#I=e^(sin(2x))(sin(2x)-1)+C#

Answer 2 · 2018-04-13T09:05:00

#= sin 2x \ e^(sin2x) - e^(sin2x) + C#

#int \ sin 4x \ e^(sin2x) \ dx#

#= 2 int \ sin 2x \cos 2x \ e^(sin2x) \ dx#

#= int \sin 2x \ d( e^(sin2x)) #

#= sin 2x \ e^(sin2x) - int \d(sin 2x) e^(sin2x) #

#= sin 2x \ e^(sin2x) - 2 int cos 2x \ e^(sin2x) \ dx#

#= sin 2x \ e^(sin2x) - int d( e^(sin2x))#

#= sin 2x \ e^(sin2x) - e^(sin2x) + C#