What is the integral of 3e^{-t}sin(t) ?

The integral of 3e^{-t}sin(t) is 3(-(e^{-t}cos(t))/2-(e^{-t}sin(t))/2)+C

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integral of 3e^{-t}sin(t)

\int 3 e^{- t} sin (t) d t

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

Solution steps

\int 3 e^{- t} sin (t) d t

Take the constant out:

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

= 3 \cdot \int e^{- t} sin (t) d t

Apply Integration By Parts

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

= 3 (- e^{- t} cos (t) - \int e^{- t} cos (t) d t)

Apply Integration By Parts

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

= 3 (- e^{- t} cos (t) - (e^{- t} sin (t) - \int - e^{- t} sin (t) d t))

Take the constant out:

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

= 3 (- e^{- t} cos (t) - (e^{- t} sin (t) - (- \int e^{- t} sin (t) d t)))

Therefore

3 \cdot \int e^{- t} sin (t) d t = 3 (- e^{- t} cos (t) - (e^{- t} sin (t) - (- \int e^{- t} sin (t) d t)))

Isolate

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

= 3 (- \frac{e ^{- t} cos ( t )}{2} - \frac{e ^{- t} sin ( t )}{2})

Add a constant to the solution

= 3 (- \frac{e ^{- t} cos ( t )}{2} - \frac{e ^{- t} sin ( t )}{2}) + C

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What is the integral of 3e^{-t}sin(t) ?
The integral of 3e^{-t}sin(t) is 3(-(e^{-t}cos(t))/2-(e^{-t}sin(t))/2)+C