What is the integral of e^xcos(2x) ?

The integral of e^xcos(2x) is (2e^xsin(2x))/5+(e^xcos(2x))/5+C

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integral of e^xcos(2x)

\int e^{x} cos (2 x) d x

\frac{2 e ^{x} sin ( 2 x )}{5} + \frac{e ^{x} cos ( 2 x )}{5} + C

Solution steps

\int e^{x} cos (2 x) d x

Apply Integration By Parts

: e^{x} \frac{1}{2} sin (2 x) - \int e^{x} \frac{1}{2} sin (2 x) d x

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

Apply Integration By Parts

: - \frac{1}{2} e^{x} cos (2 x) - \int - \frac{1}{2} e^{x} cos (2 x) d x

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

Therefore

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

Isolate

\int e^{x} cos (2 x) d x

= \frac{2 e ^{x} sin ( 2 x )}{5} + \frac{e ^{x} cos ( 2 x )}{5}

Add a constant to the solution

= \frac{2 e ^{x} sin ( 2 x )}{5} + \frac{e ^{x} cos ( 2 x )}{5} + C

\frac{d y}{d x} = k y^{2} ln (x), y (1) = - 5

(x^{3} - y^{3}) d x + x y^{2} d y = 0

sin (8 x) + sin^{2} (8 x)

3

\frac{d}{d x} (6 x cos (x))

What is the integral of e^xcos(2x) ?
The integral of e^xcos(2x) is (2e^xsin(2x))/5+(e^xcos(2x))/5+C