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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (z^{2})

x \to \frac{π}{5} lim (cot (5 x))

\frac{\partial}{\partial x} (x y - x + y)

\int \frac{x ^{2} - 16}{x ^{2}} d x

0, x^{2}, - x + 5

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