What is the integral of e^{6x}cos(2x) ?

The integral of e^{6x}cos(2x) is (e^{6x}sin(2x))/(20)+(3e^{6x}cos(2x))/(20)+C

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integral of e^{6x}cos(2x)

\int e^{6 x} cos (2 x) d x

\frac{e ^{6 x} sin ( 2 x )}{20} + \frac{3 e ^{6 x} cos ( 2 x )}{20} + C

Solution steps

\int e^{6 x} cos (2 x) d x

Apply Integration By Parts

: \frac{1}{2} e^{6 x} sin (2 x) - \int 3 e^{6 x} sin (2 x) d x

= \frac{1}{2} e^{6 x} sin (2 x) - \int 3 e^{6 x} sin (2 x) d x

Take the constant out:

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

= \frac{1}{2} e^{6 x} sin (2 x) - 3 \cdot \int e^{6 x} sin (2 x) d x

Apply Integration By Parts

: - \frac{1}{2} e^{6 x} cos (2 x) - \int - 3 e^{6 x} cos (2 x) d x

= \frac{1}{2} e^{6 x} sin (2 x) - 3 (- \frac{1}{2} e^{6 x} cos (2 x) - \int - 3 e^{6 x} cos (2 x) d x)

Take the constant out:

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

= \frac{1}{2} e^{6 x} sin (2 x) - 3 (- \frac{1}{2} e^{6 x} cos (2 x) - (- 3 \cdot \int e^{6 x} cos (2 x) d x))

Therefore

\int e^{6 x} cos (2 x) d x = \frac{1}{2} e^{6 x} sin (2 x) - 3 (- \frac{1}{2} e^{6 x} cos (2 x) - (- 3 \cdot \int e^{6 x} cos (2 x) d x))

Isolate

\int e^{6 x} cos (2 x) d x

= \frac{e ^{6 x} sin ( 2 x )}{20} + \frac{3 e ^{6 x} cos ( 2 x )}{20}

Add a constant to the solution

= \frac{e ^{6 x} sin ( 2 x )}{20} + \frac{3 e ^{6 x} cos ( 2 x )}{20} + C

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

y^{^{''}} + 2 y^{^{'}} + 1 y = 0

\frac{\partial}{\partial y} (\frac{x}{( 1 + 2 x + 3 y )})

2 y^{^{''}} - 9 y^{^{'}} + 4 y = 0

f (x) = x^{6} e^{x}

What is the integral of e^{6x}cos(2x) ?
The integral of e^{6x}cos(2x) is (e^{6x}sin(2x))/(20)+(3e^{6x}cos(2x))/(20)+C