What is the integral of 5e^{-2t}sin(t) ?

The integral of 5e^{-2t}sin(t) is -e^{-2t}cos(t)-2e^{-2t}sin(t)+C

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integral of 5e^{-2t}sin(t)

\int 5 e^{- 2 t} sin (t) d t

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

Solution steps

\int 5 e^{- 2 t} sin (t) d t

Take the constant out:

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

= 5 \cdot \int e^{- 2 t} sin (t) d t

Apply Integration By Parts

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

= 5 (- e^{- 2 t} cos (t) - \int 2 e^{- 2 t} cos (t) d t)

Take the constant out:

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

= 5 (- e^{- 2 t} cos (t) - 2 \cdot \int e^{- 2 t} cos (t) d t)

Apply Integration By Parts

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

= 5 (- e^{- 2 t} cos (t) - 2 (e^{- 2 t} sin (t) - \int - 2 e^{- 2 t} sin (t) d t))

Take the constant out:

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

= 5 (- e^{- 2 t} cos (t) - 2 (e^{- 2 t} sin (t) - (- 2 \cdot \int e^{- 2 t} sin (t) d t)))

Therefore

5 \cdot \int e^{- 2 t} sin (t) d t = 5 (- e^{- 2 t} cos (t) - 2 (e^{- 2 t} sin (t) - (- 2 \cdot \int e^{- 2 t} sin (t) d t)))

Isolate

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

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

Simplify

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

= - e^{- 2 t} cos (t) - 2 e^{- 2 t} sin (t)

Add a constant to the solution

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

\int (x + 2)^{2} (2 - x)^{6} d x

x^{3} + 3 x^{2} - 9 x

\int_{0}^{1} 1 - x^{2} d x

e^{-, at}

What is the integral of 5e^{-2t}sin(t) ?
The integral of 5e^{-2t}sin(t) is -e^{-2t}cos(t)-2e^{-2t}sin(t)+C