What is the integral of e^xcos(4x) ?

The integral of e^xcos(4x) is (4e^xsin(4x))/(17)+(e^xcos(4x))/(17)+C

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integral of e^xcos(4x)

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

\frac{4 e ^{x} sin ( 4 x )}{17} + \frac{e ^{x} cos ( 4 x )}{17} + C

Solution steps

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

Apply Integration By Parts

: e^{x} \frac{1}{4} sin (4 x) - \int e^{x} \frac{1}{4} sin (4 x) d x

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

Apply Integration By Parts

: - \frac{1}{4} e^{x} cos (4 x) - \int - \frac{1}{4} e^{x} cos (4 x) d x

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

Therefore

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

Isolate

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

= \frac{4 e ^{x} sin ( 4 x )}{17} + \frac{e ^{x} cos ( 4 x )}{17}

Add a constant to the solution

= \frac{4 e ^{x} sin ( 4 x )}{17} + \frac{e ^{x} cos ( 4 x )}{17} + C

f (x) = x^{4} - 17 x^{2} + 16, at x = - 2

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

\frac{- 5}{x ^{6}}

\int (1 + x^{2}) (2 x) d x

sec (6 x)

What is the integral of e^xcos(4x) ?
The integral of e^xcos(4x) is (4e^xsin(4x))/(17)+(e^xcos(4x))/(17)+C