What is the integral of 6e^{4x}cos(8x) ?

The integral of 6e^{4x}cos(8x) is 6((e^{4x}sin(8x))/(10)+(e^{4x}cos(8x))/(20))+C

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integral of 6e^{4x}cos(8x)

\int 6 e^{4 x} cos (8 x) d x

6 (\frac{e ^{4 x} sin ( 8 x )}{10} + \frac{e ^{4 x} cos ( 8 x )}{20}) + C

Solution steps

\int 6 e^{4 x} cos (8 x) d x

Take the constant out:

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

= 6 \cdot \int e^{4 x} cos (8 x) d x

Apply Integration By Parts

: \frac{1}{8} e^{4 x} sin (8 x) - \int \frac{1}{2} e^{4 x} sin (8 x) d x

= 6 (\frac{1}{8} e^{4 x} sin (8 x) - \int \frac{1}{2} e^{4 x} sin (8 x) d x)

Take the constant out:

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

= 6 (\frac{1}{8} e^{4 x} sin (8 x) - \frac{1}{2} \cdot \int e^{4 x} sin (8 x) d x)

Apply Integration By Parts

: - \frac{1}{8} e^{4 x} cos (8 x) - \int - \frac{1}{2} e^{4 x} cos (8 x) d x

= 6 (\frac{1}{8} e^{4 x} sin (8 x) - \frac{1}{2} (- \frac{1}{8} e^{4 x} cos (8 x) - \int - \frac{1}{2} e^{4 x} cos (8 x) d x))

Take the constant out:

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

= 6 (\frac{1}{8} e^{4 x} sin (8 x) - \frac{1}{2} (- \frac{1}{8} e^{4 x} cos (8 x) - (- \frac{1}{2} \cdot \int e^{4 x} cos (8 x) d x)))

Therefore

6 \cdot \int e^{4 x} cos (8 x) d x = 6 (\frac{1}{8} e^{4 x} sin (8 x) - \frac{1}{2} (- \frac{1}{8} e^{4 x} cos (8 x) - (- \frac{1}{2} \cdot \int e^{4 x} cos (8 x) d x)))

Isolate

\int e^{4 x} cos (8 x) d x

= 6 (\frac{e ^{4 x} sin ( 8 x )}{10} + \frac{e ^{4 x} cos ( 8 x )}{20})

Add a constant to the solution

= 6 (\frac{e ^{4 x} sin ( 8 x )}{10} + \frac{e ^{4 x} cos ( 8 x )}{20}) + C

y = \frac{1}{x - 2}, at x = 6

\frac{d}{d x} ((\frac{4}{5} x)^{5})

\int \frac{x}{( x ^{4} + 2 )} d x

t y^{^{'}} + 2 y = sin (t), y (\frac{π}{2}) = 7

x^{2} - 41, 7 - 2 x

What is the integral of 6e^{4x}cos(8x) ?
The integral of 6e^{4x}cos(8x) is 6((e^{4x}sin(8x))/(10)+(e^{4x}cos(8x))/(20))+C