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

The integral from 0 to N of e^{(-st)}sin(2t) is (-2e^{-Ns}cos(2N)-e^{-Ns}ssin(2N)+2)/(4+s^2)

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

\int_{0}^{N} e^{(- s t)} sin (2 t) d t

\frac{- 2 e ^{- N s} cos ( 2 N ) - e ^{- N s} s sin ( 2 N ) + 2}{4 + s ^{2}}

Solution steps

\int_{0}^{N} e^{- s t} sin (2 t) d t

Compute the indefinite integral:

\int e^{- s t} sin (2 t) d t = - \frac{2 e ^{- s t} cos ( 2 t )}{4 + s ^{2}} - \frac{s e ^{- s t} sin ( 2 t )}{4 + s ^{2}} + C

Compute the boundaries:

\int_{0}^{N} e^{(- s t)} sin (2 t) d t = - \frac{2 e ^{- s N} cos ( 2 N )}{4 + s ^{2}} - \frac{s e ^{- s N} sin ( 2 N )}{4 + s ^{2}} - (- \frac{2}{4 + s ^{2}})

= - \frac{2 e ^{- s N} cos ( 2 N )}{4 + s ^{2}} - \frac{s e ^{- s N} sin ( 2 N )}{4 + s ^{2}} - (- \frac{2}{4 + s ^{2}})

Simplify

= \frac{- 2 e ^{- N s} cos ( 2 N ) - e ^{- N s} s sin ( 2 N ) + 2}{4 + s ^{2}}

h \to 3 lim (\frac{3 ( x + h ) ^{3} - 3 x ^{3}}{h})

\int_{- 1}^{2} \frac{2 x - 7}{x ^{3}} d x

\frac{1}{1 - x}, 3

\int_{0}^{1} \frac{2}{x - 2} d x

\frac{d}{d x} (\frac{1}{1 + 2 x})

What is the integral from 0 to N of e^{(-st)}sin(2t) ?
The integral from 0 to N of e^{(-st)}sin(2t) is (-2e^{-Ns}cos(2N)-e^{-Ns}ssin(2N)+2)/(4+s^2)