What is the integral of 64e^{4x}sin(4x) ?

The integral of 64e^{4x}sin(4x) is 64((e^{4x}sin(4x))/8-(e^{4x}cos(4x))/8)+C

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integral of 64e^{4x}sin(4x)

\int 64 e^{4 x} sin (4 x) d x

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

Solution steps

\int 64 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

= 64 \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

= 64 (- \frac{1}{4} e^{4 x} cos (4 x) - \int - e^{4 x} cos (4 x) d x)

Take the constant out:

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

= 64 (- \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

= 64 (- \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

64 \cdot \int e^{4 x} sin (4 x) d x = 64 (- \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

= 64 (\frac{e ^{4 x} sin ( 4 x )}{8} - \frac{e ^{4 x} cos ( 4 x )}{8})

Add a constant to the solution

= 64 (\frac{e ^{4 x} sin ( 4 x )}{8} - \frac{e ^{4 x} cos ( 4 x )}{8}) + C

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

x \to 0 + lim (\frac{1}{x ln ( x )})

\int \frac{1}{x ^{6}} + \frac{3}{x ^{2}} d x

2 x e^{x} + e^{x} x^{2}

y = e^{- 7 x^{2}}

What is the integral of 64e^{4x}sin(4x) ?
The integral of 64e^{4x}sin(4x) is 64((e^{4x}sin(4x))/8-(e^{4x}cos(4x))/8)+C