What is the integral of sin(2t)e^t ?

The integral of sin(2t)e^t is -(2e^tcos(2t))/5+(e^tsin(2t))/5+C

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integral of sin(2t)e^t

\int sin (2 t) e^{t} d t

- \frac{2 e ^{t} cos ( 2 t )}{5} + \frac{e ^{t} sin ( 2 t )}{5} + C

Solution steps

\int sin (2 t) e^{t} d t

Apply Integration By Parts

= - \frac{1}{2} e^{t} cos (2 t) - \int - \frac{1}{2} e^{t} cos (2 t) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} e^{t} cos (2 t) - (- \frac{1}{2} \cdot \int e^{t} cos (2 t) d t)

Apply Integration By Parts

= - \frac{1}{2} e^{t} cos (2 t) - (- \frac{1}{2} (e^{t} \frac{1}{2} sin (2 t) - \int e^{t} \frac{1}{2} sin (2 t) d t))

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} e^{t} cos (2 t) - (- \frac{1}{2} (e^{t} \frac{1}{2} sin (2 t) - \frac{1}{2} \cdot \int e^{t} sin (2 t) d t))

Therefore

\int e^{t} sin (2 t) d t = - \frac{1}{2} e^{t} cos (2 t) - (- \frac{1}{2} (e^{t} \frac{1}{2} sin (2 t) - \frac{1}{2} \cdot \int e^{t} sin (2 t) d t))

Isolate

\int e^{t} sin (2 t) d t

= - \frac{2 e ^{t} cos ( 2 t )}{5} + \frac{e ^{t} sin ( 2 t )}{5}

Add a constant to the solution

= - \frac{2 e ^{t} cos ( 2 t )}{5} + \frac{e ^{t} sin ( 2 t )}{5} + C

What is the integral of sin(2t)e^t ?
The integral of sin(2t)e^t is -(2e^tcos(2t))/5+(e^tsin(2t))/5+C