What is the integral of e^{-st}sin(bt) ?

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

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integral of e^{-st}sin(bt)

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

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

Solution steps

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

Apply Integration By Parts

: - e^{- s t} \frac{1}{b} cos (b t) - \int s e^{- s t} \frac{1}{b} cos (b t) d t

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

Take the constant out:

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

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

Simplify

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

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

Apply Integration By Parts

: e^{- s t} \frac{1}{b} sin (b t) - \int - s e^{- s t} \frac{1}{b} sin (b t) d t

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

Take the constant out:

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

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

Simplify

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

y = 1 - x^{2}, y = x^{2} - 1

x^{3} y y^{^{'}} + x^{2} y^{2} = y^{5}

\frac{\partial}{\partial y} (x^{2} e^{- \frac{x}{y}})

\int 13 e^{u} d u

f (x) = x^{2} - 3 x + 5, at x = 2

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