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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

= - 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

= - 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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

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