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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

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