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

The integral of e^{-x}sin(4x) is -(4e^{-x}cos(4x))/(17)-(e^{-x}sin(4x))/(17)+C

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

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

- \frac{4 e ^{- x} cos ( 4 x )}{17} - \frac{e ^{- x} sin ( 4 x )}{17} + C

Solution steps

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

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

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

= - \frac{4 e ^{- x} cos ( 4 x )}{17} - \frac{e ^{- x} sin ( 4 x )}{17}

Add a constant to the solution

= - \frac{4 e ^{- x} cos ( 4 x )}{17} - \frac{e ^{- x} sin ( 4 x )}{17} + C

(x^{2} + 1) y^{^{'}} + 6 x (y - 1) = 0, y (0) = 8

\int (ln (y))^{2} d y

(x^{2} - 1) (x^{2} - 4), x = 0, x = 5

x 4 - x^{2}, 0

x + y

What is the integral of e^{-x}sin(4x) ?
The integral of e^{-x}sin(4x) is -(4e^{-x}cos(4x))/(17)-(e^{-x}sin(4x))/(17)+C