What is the integral from 0 to 2pi of e^{-st}sin(t) ?

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

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integral from 0 to 2pi of e^{-st}sin(t)

\int_{0}^{2 π} e^{- s t} sin (t) d t

\frac{- e ^{- 2 π s} + 1}{s ^{2} + 1}

Solution steps

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

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

Simplify

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

\int_{- 2}^{2} (1 + 4 - x^{2}) d x

\int_{0}^{\frac{π}{3}} 2 cos (x) + 3 d x

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

\int_{0}^{1} (x^{2} + 3 x - 1) e^{x} d x

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