What is the integral of e^xcos(6x) ?

The integral of e^xcos(6x) is (6e^xsin(6x))/(37)+(e^xcos(6x))/(37)+C

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integral of e^xcos(6x)

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

\frac{6 e ^{x} sin ( 6 x )}{37} + \frac{e ^{x} cos ( 6 x )}{37} + C

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

= \frac{6 e ^{x} sin ( 6 x )}{37} + \frac{e ^{x} cos ( 6 x )}{37}

Add a constant to the solution

= \frac{6 e ^{x} sin ( 6 x )}{37} + \frac{e ^{x} cos ( 6 x )}{37} + C

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What is the integral of e^xcos(6x) ?
The integral of e^xcos(6x) is (6e^xsin(6x))/(37)+(e^xcos(6x))/(37)+C