What is the integral of e^{-sx}cos(8x) ?

The integral of e^{-sx}cos(8x) is (8e^{-sx}sin(8x))/(64+s^2)-(se^{-sx}cos(8x))/(64+s^2)+C

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integral of e^{-sx}cos(8x)

\int e^{- s x} cos (8 x) d x

\frac{8 e ^{- s x} sin ( 8 x )}{64 + s ^{2}} - \frac{s e ^{- s x} cos ( 8 x )}{64 + s ^{2}} + C

Solution steps

\int e^{- s x} cos (8 x) d x

Apply Integration By Parts

: \frac{1}{8} e^{- s x} sin (8 x) - \int - \frac{1}{8} s e^{- s x} sin (8 x) d x

= \frac{1}{8} e^{- s x} sin (8 x) - \int - \frac{1}{8} s e^{- s x} sin (8 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} e^{- s x} sin (8 x) - (- \frac{1}{8} s \cdot \int e^{- s x} sin (8 x) d x)

Simplify

- \frac{1}{8} s : - \frac{s}{8}

= \frac{1}{8} e^{- s x} sin (8 x) - (- \frac{s}{8} \cdot \int e^{- s x} sin (8 x) d x)

Apply Integration By Parts

: - \frac{1}{8} e^{- s x} cos (8 x) - \int \frac{1}{8} s e^{- s x} cos (8 x) d x

= \frac{1}{8} e^{- s x} sin (8 x) - (- \frac{s}{8} (- \frac{1}{8} e^{- s x} cos (8 x) - \int \frac{1}{8} s e^{- s x} cos (8 x) d x))

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} e^{- s x} sin (8 x) - (- \frac{s}{8} (- \frac{1}{8} e^{- s x} cos (8 x) - \frac{1}{8} s \cdot \int e^{- s x} cos (8 x) d x))

Simplify

\frac{1}{8} s : \frac{s}{8}

= \frac{1}{8} e^{- s x} sin (8 x) - (- \frac{s}{8} (- \frac{1}{8} e^{- s x} cos (8 x) - \frac{s}{8} \cdot \int e^{- s x} cos (8 x) d x))

Therefore

\int e^{- s x} cos (8 x) d x = \frac{1}{8} e^{- s x} sin (8 x) - (- \frac{s}{8} (- \frac{1}{8} e^{- s x} cos (8 x) - \frac{s}{8} \cdot \int e^{- s x} cos (8 x) d x))

Isolate

\int e^{- s x} cos (8 x) d x

= \frac{8 e ^{- s x} sin ( 8 x )}{64 + s ^{2}} - \frac{s e ^{- s x} cos ( 8 x )}{64 + s ^{2}}

Add a constant to the solution

= \frac{8 e ^{- s x} sin ( 8 x )}{64 + s ^{2}} - \frac{s e ^{- s x} cos ( 8 x )}{64 + s ^{2}} + C

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What is the integral of e^{-sx}cos(8x) ?
The integral of e^{-sx}cos(8x) is (8e^{-sx}sin(8x))/(64+s^2)-(se^{-sx}cos(8x))/(64+s^2)+C