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 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

By integration by parts

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

Apr 13, 2018

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

Explanation:

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