What is the integral from 0 to t of e^{-2u}sin(u) ?

The integral from 0 to t of e^{-2u}sin(u) is (-e^{-2t}cos(t)-2e^{-2t}sin(t)+1)/5

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integral from 0 to t of e^{-2u}sin(u)

\int_{0}^{t} e^{- 2 u} sin (u) d u

\frac{- e ^{- 2 t} cos ( t ) - 2 e ^{- 2 t} sin ( t ) + 1}{5}

Solution steps

\int_{0}^{t} e^{- 2 u} sin (u) d u

Compute the indefinite integral:

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

Compute the boundaries:

\int_{0}^{t} e^{- 2 u} sin (u) d u = - \frac{e ^{- 2 t} cos ( t )}{5} - \frac{2 e ^{- 2 t} sin ( t )}{5} - (- \frac{1}{5})

= - \frac{e ^{- 2 t} cos ( t )}{5} - \frac{2 e ^{- 2 t} sin ( t )}{5} - (- \frac{1}{5})

Simplify

= \frac{- e ^{- 2 t} cos ( t ) - 2 e ^{- 2 t} sin ( t ) + 1}{5}

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What is the integral from 0 to t of e^{-2u}sin(u) ?
The integral from 0 to t of e^{-2u}sin(u) is (-e^{-2t}cos(t)-2e^{-2t}sin(t)+1)/5