What is the integral of 16e^{-4x}sin(4x) ?

The integral of 16e^{-4x}sin(4x) is 16(-(e^{-4x}cos(4x))/8-(e^{-4x}sin(4x))/8)+C

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integral of 16e^{-4x}sin(4x)

\int 16 e^{- 4 x} sin (4 x) d x

16 (- \frac{e ^{- 4 x} cos ( 4 x )}{8} - \frac{e ^{- 4 x} sin ( 4 x )}{8}) + C

Solution steps

\int 16 e^{- 4 x} sin (4 x) d x

Take the constant out:

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

= 16 \cdot \int e^{- 4 x} sin (4 x) d x

Apply Integration By Parts

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

= 16 (- \frac{1}{4} e^{- 4 x} cos (4 x) - \int e^{- 4 x} cos (4 x) d x)

Apply Integration By Parts

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

= 16 (- \frac{1}{4} e^{- 4 x} cos (4 x) - (\frac{1}{4} e^{- 4 x} sin (4 x) - \int - e^{- 4 x} sin (4 x) d x))

Take the constant out:

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

= 16 (- \frac{1}{4} e^{- 4 x} cos (4 x) - (\frac{1}{4} e^{- 4 x} sin (4 x) - (- \int e^{- 4 x} sin (4 x) d x)))

Therefore

16 \cdot \int e^{- 4 x} sin (4 x) d x = 16 (- \frac{1}{4} e^{- 4 x} cos (4 x) - (\frac{1}{4} e^{- 4 x} sin (4 x) - (- \int e^{- 4 x} sin (4 x) d x)))

Isolate

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

= 16 (- \frac{e ^{- 4 x} cos ( 4 x )}{8} - \frac{e ^{- 4 x} sin ( 4 x )}{8})

Add a constant to the solution

= 16 (- \frac{e ^{- 4 x} cos ( 4 x )}{8} - \frac{e ^{- 4 x} sin ( 4 x )}{8}) + C

- 6 s

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

\frac{1}{2 + x}, 0

y^{^{''}} + y = 5 sin (x)

\frac{d y}{d x} x - x e^{x} + y = 0

What is the integral of 16e^{-4x}sin(4x) ?
The integral of 16e^{-4x}sin(4x) is 16(-(e^{-4x}cos(4x))/8-(e^{-4x}sin(4x))/8)+C