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

The integral of sin(x)e^{-sx} is -(e^{-sx}cos(x))/(1+s^2)-(se^{-sx}sin(x))/(1+s^2)+C

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integral of sin(x)e^{-sx}

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

- \frac{e ^{- s x} cos ( x )}{1 + s ^{2}} - \frac{s e ^{- s x} sin ( x )}{1 + s ^{2}} + C

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

= - \frac{e ^{- s x} cos ( x )}{1 + s ^{2}} - \frac{s e ^{- s x} sin ( x )}{1 + s ^{2}}

Add a constant to the solution

= - \frac{e ^{- s x} cos ( x )}{1 + s ^{2}} - \frac{s e ^{- s x} sin ( x )}{1 + s ^{2}} + C

\int \frac{1}{v ^{2} - v + 1} d v

f (x) = ln (\frac{x + 1}{x - 1})

\int \frac{1}{x ^{2} 1 - x ^{2}} d x

5 x^{2} + 4 x + 2

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (a x)

What is the integral of sin(x)e^{-sx} ?
The integral of sin(x)e^{-sx} is -(e^{-sx}cos(x))/(1+s^2)-(se^{-sx}sin(x))/(1+s^2)+C