What is the integral of e^{3x}cos(6x) ?

The integral of e^{3x}cos(6x) is (2e^{3x}sin(6x))/(15)+(e^{3x}cos(6x))/(15)+C

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integral of e^{3x}cos(6x)

\int e^{3 x} cos (6 x) d x

\frac{2 e ^{3 x} sin ( 6 x )}{15} + \frac{e ^{3 x} cos ( 6 x )}{15} + C

Solution steps

\int e^{3 x} cos (6 x) d x

Apply Integration By Parts

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

Apply Integration By Parts

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

Therefore

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

Isolate

\int e^{3 x} cos (6 x) d x

= \frac{2 e ^{3 x} sin ( 6 x )}{15} + \frac{e ^{3 x} cos ( 6 x )}{15}

Add a constant to the solution

= \frac{2 e ^{3 x} sin ( 6 x )}{15} + \frac{e ^{3 x} cos ( 6 x )}{15} + C

What is the integral of e^{3x}cos(6x) ?
The integral of e^{3x}cos(6x) is (2e^{3x}sin(6x))/(15)+(e^{3x}cos(6x))/(15)+C