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

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

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

\int_{0}^{π} sin (t) e^{- s t} d t

\frac{e ^{- π s} + 1}{s ^{2} + 1}

Solution steps

\int_{0}^{π} sin (t) e^{- s t} d t

Compute the indefinite integral:

\int sin (t) e^{- s 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} sin (t) e^{- s 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} (sec (x) - 5 x^{4})

x \to - 6 lim (\frac{∣ x + 6 ∣}{x + 6})

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

\int tan^{7} (x) sec^{3} (x) d x

\frac{d}{d x} (\frac{sin ( x )}{x ^{2} - 1})

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