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

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

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

\int_{0}^{π} e^{- s t} sin (t) d t

\frac{e ^{- π s} + 1}{s ^{2} + 1}

Solution steps

\int_{0}^{π} e^{- s t} sin (t) d t

Compute the indefinite integral:

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

Compute the boundaries:

\int_{0}^{π} e^{- s t} sin (t) d t = \frac{e ^{- π s}}{s ^{2} + 1} - (- \frac{1}{1 + s ^{2}})

= \frac{e ^{- π s}}{s ^{2} + 1} - (- \frac{1}{1 + s ^{2}})

Simplify

= \frac{e ^{- π s} + 1}{s ^{2} + 1}

\frac{d}{d x} (\frac{cos ( x )}{2 + sin ( x )})

\frac{d y}{d x} = e^{4 x + 2 y}

\int ((x^{2} + 7)^{2021}) x d x

\int_{1}^{2} \frac{x ^{2} + 1}{3 x - x ^{2}} d x

\frac{d}{d x} ((x + (x + sin^{2} (x))^{3})^{4})

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