What is the integral of e^{-5x}sin(4x) ?

The integral of e^{-5x}sin(4x) is -(4e^{-5x}cos(4x))/(41)-(5e^{-5x}sin(4x))/(41)+C

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integral of e^{-5x}sin(4x)

\int e^{- 5 x} sin (4 x) d x

- \frac{4 e ^{- 5 x} cos ( 4 x )}{41} - \frac{5 e ^{- 5 x} sin ( 4 x )}{41} + C

Solution steps

\int e^{- 5 x} sin (4 x) d x

Apply Integration By Parts

: - \frac{1}{4} e^{- 5 x} cos (4 x) - \int \frac{5}{4} e^{- 5 x} cos (4 x) d x

= - \frac{1}{4} e^{- 5 x} cos (4 x) - \int \frac{5}{4} e^{- 5 x} cos (4 x) d x

Take the constant out:

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

= - \frac{1}{4} e^{- 5 x} cos (4 x) - \frac{5}{4} \cdot \int e^{- 5 x} cos (4 x) d x

Apply Integration By Parts

: \frac{1}{4} e^{- 5 x} sin (4 x) - \int - \frac{5}{4} e^{- 5 x} sin (4 x) d x

= - \frac{1}{4} e^{- 5 x} cos (4 x) - \frac{5}{4} (\frac{1}{4} e^{- 5 x} sin (4 x) - \int - \frac{5}{4} e^{- 5 x} sin (4 x) d x)

Take the constant out:

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

= - \frac{1}{4} e^{- 5 x} cos (4 x) - \frac{5}{4} (\frac{1}{4} e^{- 5 x} sin (4 x) - (- \frac{5}{4} \cdot \int e^{- 5 x} sin (4 x) d x))

Therefore

\int e^{- 5 x} sin (4 x) d x = - \frac{1}{4} e^{- 5 x} cos (4 x) - \frac{5}{4} (\frac{1}{4} e^{- 5 x} sin (4 x) - (- \frac{5}{4} \cdot \int e^{- 5 x} sin (4 x) d x))

Isolate

\int e^{- 5 x} sin (4 x) d x

= - \frac{4 e ^{- 5 x} cos ( 4 x )}{41} - \frac{5 e ^{- 5 x} sin ( 4 x )}{41}

Add a constant to the solution

= - \frac{4 e ^{- 5 x} cos ( 4 x )}{41} - \frac{5 e ^{- 5 x} sin ( 4 x )}{41} + C

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

x^{2} y^{^{''}} - 3 x y^{^{'}} + 4 y = 0

ln (x), x = 4

y^{^{''}} + y = sin (x), y (0) = 1, y (\frac{π}{2}) = 0

x^{7} (1 - \frac{2}{x + 4})

What is the integral of e^{-5x}sin(4x) ?
The integral of e^{-5x}sin(4x) is -(4e^{-5x}cos(4x))/(41)-(5e^{-5x}sin(4x))/(41)+C