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

The integral of e^{4x}cos(6x) is (3e^{4x}sin(6x))/(26)+(e^{4x}cos(6x))/(13)+C

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

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

\frac{3 e ^{4 x} sin ( 6 x )}{26} + \frac{e ^{4 x} cos ( 6 x )}{13} + C

Solution steps

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

Apply Integration By Parts

= \frac{1}{6} e^{4 x} sin (6 x) - \int \frac{2}{3} e^{4 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^{4 x} sin (6 x) - \frac{2}{3} \cdot \int e^{4 x} sin (6 x) d x

Apply Integration By Parts

= \frac{1}{6} e^{4 x} sin (6 x) - \frac{2}{3} (- \frac{1}{6} e^{4 x} cos (6 x) - \int - \frac{2}{3} e^{4 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^{4 x} sin (6 x) - \frac{2}{3} (- \frac{1}{6} e^{4 x} cos (6 x) - (- \frac{2}{3} \cdot \int e^{4 x} cos (6 x) d x))

Therefore

\int e^{4 x} cos (6 x) d x = \frac{1}{6} e^{4 x} sin (6 x) - \frac{2}{3} (- \frac{1}{6} e^{4 x} cos (6 x) - (- \frac{2}{3} \cdot \int e^{4 x} cos (6 x) d x))

Isolate

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

= \frac{3 e ^{4 x} sin ( 6 x )}{26} + \frac{e ^{4 x} cos ( 6 x )}{13}

Add a constant to the solution

= \frac{3 e ^{4 x} sin ( 6 x )}{26} + \frac{e ^{4 x} cos ( 6 x )}{13} + C

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What is the integral of e^{4x}cos(6x) ?
The integral of e^{4x}cos(6x) is (3e^{4x}sin(6x))/(26)+(e^{4x}cos(6x))/(13)+C