What is the integral of 64e^{-8x}sin(8x) ?

The integral of 64e^{-8x}sin(8x) is 64(-(e^{-8x}cos(8x))/(16)-(e^{-8x}sin(8x))/(16))+C

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integral of 64e^{-8x}sin(8x)

\int 64 e^{- 8 x} sin (8 x) d x

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

Solution steps

\int 64 e^{- 8 x} sin (8 x) d x

Take the constant out:

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

= 64 \cdot \int e^{- 8 x} sin (8 x) d x

Apply Integration By Parts

: - \frac{1}{8} e^{- 8 x} cos (8 x) - \int e^{- 8 x} cos (8 x) d x

= 64 (- \frac{1}{8} e^{- 8 x} cos (8 x) - \int e^{- 8 x} cos (8 x) d x)

Apply Integration By Parts

: \frac{1}{8} e^{- 8 x} sin (8 x) - \int - e^{- 8 x} sin (8 x) d x

= 64 (- \frac{1}{8} e^{- 8 x} cos (8 x) - (\frac{1}{8} e^{- 8 x} sin (8 x) - \int - e^{- 8 x} sin (8 x) d x))

Take the constant out:

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

= 64 (- \frac{1}{8} e^{- 8 x} cos (8 x) - (\frac{1}{8} e^{- 8 x} sin (8 x) - (- \int e^{- 8 x} sin (8 x) d x)))

Therefore

64 \cdot \int e^{- 8 x} sin (8 x) d x = 64 (- \frac{1}{8} e^{- 8 x} cos (8 x) - (\frac{1}{8} e^{- 8 x} sin (8 x) - (- \int e^{- 8 x} sin (8 x) d x)))

Isolate

\int e^{- 8 x} sin (8 x) d x

= 64 (- \frac{e ^{- 8 x} cos ( 8 x )}{16} - \frac{e ^{- 8 x} sin ( 8 x )}{16})

Add a constant to the solution

= 64 (- \frac{e ^{- 8 x} cos ( 8 x )}{16} - \frac{e ^{- 8 x} sin ( 8 x )}{16}) + C

\frac{d}{d x} (ln (x) + x)

\frac{d}{d x} (- 2 x + y)

\int u + 1 d u

y = sec (1 + 3 4 + arcsin (4 + π^{2}))

10 (t + 1) \frac{d y}{d t} - 9 y = 9 t

What is the integral of 64e^{-8x}sin(8x) ?
The integral of 64e^{-8x}sin(8x) is 64(-(e^{-8x}cos(8x))/(16)-(e^{-8x}sin(8x))/(16))+C