What is the integral of-1/2*e^{-2t}*sin(2t) ?

The integral of-1/2*e^{-2t}*sin(2t) is -1/8 (-e^{-2t}cos(2t)-e^{-2t}sin(2t))+C

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integral of-1/2e^{-2t}sin(2t)

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

- \frac{1}{8} (- e^{- 2 t} cos (2 t) - e^{- 2 t} sin (2 t)) + C

Solution steps

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

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

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

= - \frac{1}{2} (- \frac{e ^{- 2 t} cos ( 2 t )}{4} - \frac{e ^{- 2 t} sin ( 2 t )}{4})

Simplify

- \frac{1}{2} (- \frac{e ^{- 2 t} cos ( 2 t )}{4} - \frac{e ^{- 2 t} sin ( 2 t )}{4}) : - \frac{1}{8} (- e^{- 2 t} cos (2 t) - e^{- 2 t} sin (2 t))

= - \frac{1}{8} (- e^{- 2 t} cos (2 t) - e^{- 2 t} sin (2 t))

Add a constant to the solution

= - \frac{1}{8} (- e^{- 2 t} cos (2 t) - e^{- 2 t} sin (2 t)) + C

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What is the integral of-1/2*e^{-2t}*sin(2t) ?
The integral of-1/2*e^{-2t}*sin(2t) is -1/8 (-e^{-2t}cos(2t)-e^{-2t}sin(2t))+C