What is the integral from 0 to N of e^{(-st)}sin(5t) ?

The integral from 0 to N of e^{(-st)}sin(5t) is (-5e^{-Ns}cos(5N)-e^{-Ns}ssin(5N)+5)/(25+s^2)

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integral from 0 to N of e^{(-st)}sin(5t)

\int_{0}^{N} e^{(- s t)} sin (5 t) d t

\frac{- 5 e ^{- N s} cos ( 5 N ) - e ^{- N s} s sin ( 5 N ) + 5}{25 + s ^{2}}

Solution steps

\int_{0}^{N} e^{- s t} sin (5 t) d t

Compute the indefinite integral:

\int e^{- s t} sin (5 t) d t = - \frac{5 e ^{- s t} cos ( 5 t )}{25 + s ^{2}} - \frac{s e ^{- s t} sin ( 5 t )}{25 + s ^{2}} + C

Compute the boundaries:

\int_{0}^{N} e^{(- s t)} sin (5 t) d t = - \frac{5 e ^{- s N} cos ( 5 N )}{25 + s ^{2}} - \frac{s e ^{- s N} sin ( 5 N )}{25 + s ^{2}} - (- \frac{5}{25 + s ^{2}})

= - \frac{5 e ^{- s N} cos ( 5 N )}{25 + s ^{2}} - \frac{s e ^{- s N} sin ( 5 N )}{25 + s ^{2}} - (- \frac{5}{25 + s ^{2}})

Simplify

= \frac{- 5 e ^{- N s} cos ( 5 N ) - e ^{- N s} s sin ( 5 N ) + 5}{25 + s ^{2}}

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

\int \frac{1}{x - 1} d x

x \to 2 lim (2 - x^{2})

\int_{1}^{e^{4}} x^{4} ln (x) d x

f (x) = x^{2} - 2 x + 5, at x = 1

What is the integral from 0 to N of e^{(-st)}sin(5t) ?
The integral from 0 to N of e^{(-st)}sin(5t) is (-5e^{-Ns}cos(5N)-e^{-Ns}ssin(5N)+5)/(25+s^2)