How do you find the indefinite integral of ∫sin 4xe^sin2x dx?
2 Answers
Apr 13, 2018
I=e^(sin(2x))(sin(2x)-1)+C
Explanation:
We want to solve
I=intsin(4x)e^(sin(2x))dx
Using the trig identity
I=2intcos(2x)sin(2x)e^(sin(2x))dx
Make a substitution
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
I=e^(sin(2x))(sin(2x)-1)+C
Apr 13, 2018