What is the integral of e^{-9u}sin(4u) ?

The integral of e^{-9u}sin(4u) is -(4e^{-9u}cos(4u))/(97)-(9e^{-9u}sin(4u))/(97)+C

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integral of e^{-9u}sin(4u)

\int e^{- 9 u} sin (4 u) d u

- \frac{4 e ^{- 9 u} cos ( 4 u )}{97} - \frac{9 e ^{- 9 u} sin ( 4 u )}{97} + C

Solution steps

\int e^{- 9 u} sin (4 u) d u

Apply Integration By Parts

: - \frac{1}{4} e^{- 9 u} cos (4 u) - \int \frac{9}{4} e^{- 9 u} cos (4 u) d u

= - \frac{1}{4} e^{- 9 u} cos (4 u) - \int \frac{9}{4} e^{- 9 u} cos (4 u) d u

Take the constant out:

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

= - \frac{1}{4} e^{- 9 u} cos (4 u) - \frac{9}{4} \cdot \int e^{- 9 u} cos (4 u) d u

Apply Integration By Parts

: \frac{1}{4} e^{- 9 u} sin (4 u) - \int - \frac{9}{4} e^{- 9 u} sin (4 u) d u

= - \frac{1}{4} e^{- 9 u} cos (4 u) - \frac{9}{4} (\frac{1}{4} e^{- 9 u} sin (4 u) - \int - \frac{9}{4} e^{- 9 u} sin (4 u) d u)

Take the constant out:

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

= - \frac{1}{4} e^{- 9 u} cos (4 u) - \frac{9}{4} (\frac{1}{4} e^{- 9 u} sin (4 u) - (- \frac{9}{4} \cdot \int e^{- 9 u} sin (4 u) d u))

Therefore

\int e^{- 9 u} sin (4 u) d u = - \frac{1}{4} e^{- 9 u} cos (4 u) - \frac{9}{4} (\frac{1}{4} e^{- 9 u} sin (4 u) - (- \frac{9}{4} \cdot \int e^{- 9 u} sin (4 u) d u))

Isolate

\int e^{- 9 u} sin (4 u) d u

= - \frac{4 e ^{- 9 u} cos ( 4 u )}{97} - \frac{9 e ^{- 9 u} sin ( 4 u )}{97}

Add a constant to the solution

= - \frac{4 e ^{- 9 u} cos ( 4 u )}{97} - \frac{9 e ^{- 9 u} sin ( 4 u )}{97} + C

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What is the integral of e^{-9u}sin(4u) ?
The integral of e^{-9u}sin(4u) is -(4e^{-9u}cos(4u))/(97)-(9e^{-9u}sin(4u))/(97)+C