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

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

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

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

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

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

\int \frac{4}{( x - 6 ) ^{3}} d x

100 x + \frac{d y}{d x} = x y

t \to 10 lim (\frac{t ^{2} + 3 t - 130}{t ^{2} - 100})

\frac{d}{d x} (\frac{e ^{- x}}{1 + x ^{2}})

\int \frac{sec ( x )}{( sec ( x ) - tan ( x ) )} d x

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