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

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

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Answer 1 · 2018-04-13T09:03:50

$I = e^{sin (2 x)} (sin (2 x) - 1) + C$ $I=e^(sin(2x))(sin(2x)-1)+C$

Explanation:

We want to solve

$I = \int sin (4 x) e^{sin (2 x)} d x$ $I=intsin(4x)e^(sin(2x))dx$

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

$I = 2 \int cos (2 x) sin (2 x) e^{sin (2 x)} d x$ $I=2intcos(2x)sin(2x)e^(sin(2x))dx$

Make a substitution $u = sin (2 x) \Rightarrow d u = 2 cos (2 x) d x$ $u=sin(2x)=>du=2cos(2x)dx$

$I = \int u e^{u} d u$ $I=intue^(u)du$

By integration by parts

$I = u e^{u} - \int e^{u} d u$ $I=ue^u-inte^udu$

$I = u e^{u} - e^{u} + C$ $color(white)(I)=ue^u-e^u+C$

$I = e^{u} (u - 1) + C$ $color(white)(I)=e^u(u-1)+C$

Substitute back $u = sin (2 x)$ $u=sin(2x)$

$I = e^{sin (2 x)} (sin (2 x) - 1) + C$ $I=e^(sin(2x))(sin(2x)-1)+C$

Answer 2 · 2018-04-13T09:05:00

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

Explanation:

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

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

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

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

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

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

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

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

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