What is the integral from 0 to N of e^{-st}sin(3t) ?

The integral from 0 to N of e^{-st}sin(3t) is (-3e^{-Ns}cos(3N)-e^{-Ns}ssin(3N)+3)/(9+s^2)

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integral from 0 to N of e^{-st}sin(3t)

\int_{0}^{N} e^{- s t} sin (3 t) d t

\frac{- 3 e ^{- N s} cos ( 3 N ) - e ^{- N s} s sin ( 3 N ) + 3}{9 + s ^{2}}

Solution steps

\int_{0}^{N} e^{- s t} sin (3 t) d t

Compute the indefinite integral:

\int e^{- s t} sin (3 t) d t = - \frac{3 e ^{- s t} cos ( 3 t )}{9 + s ^{2}} - \frac{s e ^{- s t} sin ( 3 t )}{9 + s ^{2}} + C

Compute the boundaries:

\int_{0}^{N} e^{- s t} sin (3 t) d t = - \frac{3 e ^{- s N} cos ( 3 N )}{9 + s ^{2}} - \frac{s e ^{- s N} sin ( 3 N )}{9 + s ^{2}} - (- \frac{3}{9 + s ^{2}})

= - \frac{3 e ^{- s N} cos ( 3 N )}{9 + s ^{2}} - \frac{s e ^{- s N} sin ( 3 N )}{9 + s ^{2}} - (- \frac{3}{9 + s ^{2}})

Simplify

= \frac{- 3 e ^{- N s} cos ( 3 N ) - e ^{- N s} s sin ( 3 N ) + 3}{9 + s ^{2}}

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

2 y^{^{'}} = y^{3} cos (x)

x \to 0 lim (x^{4} sin (\frac{2}{x}))

g (u) = 7 u + 13 u

\frac{\partial}{\partial x} (x e^{3 x y})

What is the integral from 0 to N of e^{-st}sin(3t) ?
The integral from 0 to N of e^{-st}sin(3t) is (-3e^{-Ns}cos(3N)-e^{-Ns}ssin(3N)+3)/(9+s^2)