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

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

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

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

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

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

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