What is the integral from 0 to N of e^{-st}sin(4t) ?

The integral from 0 to N of e^{-st}sin(4t) is (-4e^{-Ns}cos(4N)-e^{-Ns}ssin(4N)+4)/(16+s^2)

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integral from 0 to N of e^{-st}sin(4t)

\int_{0}^{N} e^{- s t} sin (4 t) d t

\frac{- 4 e ^{- N s} cos ( 4 N ) - e ^{- N s} s sin ( 4 N ) + 4}{16 + s ^{2}}

Solution steps

\int_{0}^{N} e^{- s t} sin (4 t) d t

Compute the indefinite integral:

\int e^{- s t} sin (4 t) d t = - \frac{4 e ^{- s t} cos ( 4 t )}{16 + s ^{2}} - \frac{s e ^{- s t} sin ( 4 t )}{16 + s ^{2}} + C

Compute the boundaries:

\int_{0}^{N} e^{- s t} sin (4 t) d t = - \frac{4 e ^{- s N} cos ( 4 N )}{16 + s ^{2}} - \frac{s e ^{- s N} sin ( 4 N )}{16 + s ^{2}} - (- \frac{4}{16 + s ^{2}})

= - \frac{4 e ^{- s N} cos ( 4 N )}{16 + s ^{2}} - \frac{s e ^{- s N} sin ( 4 N )}{16 + s ^{2}} - (- \frac{4}{16 + s ^{2}})

Simplify

= \frac{- 4 e ^{- N s} cos ( 4 N ) - e ^{- N s} s sin ( 4 N ) + 4}{16 + s ^{2}}

x \to - 2 lim (- 2 x^{2} + 4 x - 9)

y^{^{'}} = \frac{2 x ^{3}}{y ^{2}}, y (2) = 1

\int_{0}^{5} π y^{3} d y

\frac{\partial}{\partial x} (\frac{200}{( x ^{2} + y ^{2} )})

\frac{2}{2 s + 1}

What is the integral from 0 to N of e^{-st}sin(4t) ?
The integral from 0 to N of e^{-st}sin(4t) is (-4e^{-Ns}cos(4N)-e^{-Ns}ssin(4N)+4)/(16+s^2)