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}}

\frac{1}{1 + 4 t ^{2}}

\frac{4 x}{x + 1} (3.3)

y = t (ln (3 t))^{2}

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

\frac{5}{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)