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

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

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

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

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

Solution steps

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

\int 4 + x^{2} d x

x \to 3 lim (\frac{1 - x}{x ^{2} - 9})

d y - (y - 9)^{2} \cdot d x = 0

\frac{d}{d x} (\frac{cos ( m x )}{sin ( n x )})

\int_{0}^{4} (9 e^{- 2 x}) d x

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