What is the integral of e^{-v}sin(wv) ?

The integral of e^{-v}sin(wv) is -(we^{-v}cos(wv))/(w^2+1)-(e^{-v}sin(wv))/(w^2+1)+C

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integral of e^{-v}sin(wv)

\int e^{- v} sin (w v) d v

- \frac{w e ^{- v} cos ( w v )}{w ^{2} + 1} - \frac{e ^{- v} sin ( w v )}{w ^{2} + 1} + C

Solution steps

\int e^{- v} sin (w v) d v

Apply Integration By Parts

: - e^{- v} \frac{1}{w} cos (w v) - \int e^{- v} \frac{1}{w} cos (w v) d v

= - e^{- v} \frac{1}{w} cos (w v) - \int e^{- v} \frac{1}{w} cos (w v) d v

Take the constant out:

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

= - e^{- v} \frac{1}{w} cos (w v) - \frac{1}{w} \cdot \int e^{- v} cos (w v) d v

Apply Integration By Parts

: e^{- v} \frac{1}{w} sin (w v) - \int - e^{- v} \frac{1}{w} sin (w v) d v

= - e^{- v} \frac{1}{w} cos (w v) - \frac{1}{w} (e^{- v} \frac{1}{w} sin (w v) - \int - e^{- v} \frac{1}{w} sin (w v) d v)

Take the constant out:

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

= - e^{- v} \frac{1}{w} cos (w v) - \frac{1}{w} (e^{- v} \frac{1}{w} sin (w v) - (- \frac{1}{w} \cdot \int e^{- v} sin (w v) d v))

Therefore

\int e^{- v} sin (w v) d v = - e^{- v} \frac{1}{w} cos (w v) - \frac{1}{w} (e^{- v} \frac{1}{w} sin (w v) - (- \frac{1}{w} \cdot \int e^{- v} sin (w v) d v))

Isolate

\int e^{- v} sin (w v) d v

= - \frac{w e ^{- v} cos ( w v )}{w ^{2} + 1} - \frac{e ^{- v} sin ( w v )}{w ^{2} + 1}

Add a constant to the solution

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

\frac{d y}{d t} = - \frac{t}{y}, y (0) = 4

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

x^{4} + 6 x^{2} - x

\int \frac{sin ( y )}{cos ( y )} d y

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

What is the integral of e^{-v}sin(wv) ?
The integral of e^{-v}sin(wv) is -(we^{-v}cos(wv))/(w^2+1)-(e^{-v}sin(wv))/(w^2+1)+C