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

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

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

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

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

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

\int_{100}^{t} sec (2 x - 7) d x

\int_{0}^{1} \frac{1}{x ^{0.99}} d x

\int_{0}^{2} \frac{x}{1 + x ^{4}} d x

\int_{4}^{12} \frac{1}{8} d x

\int_{0}^{1} (6 - 9 x)^{2} d x

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