What is the integral of e^{-sx}*cos(bx) ?

The integral of e^{-sx}*cos(bx) is (be^{-sx}sin(bx))/(b^2+s^2)-(se^{-sx}cos(bx))/(b^2+s^2)+C

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integral of e^{-sx}*cos(bx)

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

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

Solution steps

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

Apply Integration By Parts

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

Take the constant out:

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

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

Simplify

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

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

Apply Integration By Parts

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

Take the constant out:

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

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

Simplify

s \frac{1}{b} : \frac{s}{b}

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

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What is the integral of e^{-sx}*cos(bx) ?
The integral of e^{-sx}*cos(bx) is (be^{-sx}sin(bx))/(b^2+s^2)-(se^{-sx}cos(bx))/(b^2+s^2)+C