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

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

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

\int_{0}^{π} sin (x) e^{- s x} d x

\frac{e ^{- π s} + 1}{s ^{2} + 1}

Solution steps

\int_{0}^{π} sin (x) e^{- s x} d x

Compute the indefinite integral:

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

Compute the boundaries:

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

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