What is the integral of e^{2x}cos(5x) ?

The integral of e^{2x}cos(5x) is (5e^{2x}sin(5x))/(29)+(2e^{2x}cos(5x))/(29)+C

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integral of e^{2x}cos(5x)

\int e^{2 x} cos (5 x) d x

\frac{5 e ^{2 x} sin ( 5 x )}{29} + \frac{2 e ^{2 x} cos ( 5 x )}{29} + C

Solution steps

\int e^{2 x} cos (5 x) d x

Apply Integration By Parts

: \frac{1}{5} e^{2 x} sin (5 x) - \int \frac{2}{5} e^{2 x} sin (5 x) d x

= \frac{1}{5} e^{2 x} sin (5 x) - \int \frac{2}{5} e^{2 x} sin (5 x) d x

Take the constant out:

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

= \frac{1}{5} e^{2 x} sin (5 x) - \frac{2}{5} \cdot \int e^{2 x} sin (5 x) d x

Apply Integration By Parts

: - \frac{1}{5} e^{2 x} cos (5 x) - \int - \frac{2}{5} e^{2 x} cos (5 x) d x

= \frac{1}{5} e^{2 x} sin (5 x) - \frac{2}{5} (- \frac{1}{5} e^{2 x} cos (5 x) - \int - \frac{2}{5} e^{2 x} cos (5 x) d x)

Take the constant out:

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

= \frac{1}{5} e^{2 x} sin (5 x) - \frac{2}{5} (- \frac{1}{5} e^{2 x} cos (5 x) - (- \frac{2}{5} \cdot \int e^{2 x} cos (5 x) d x))

Therefore

\int e^{2 x} cos (5 x) d x = \frac{1}{5} e^{2 x} sin (5 x) - \frac{2}{5} (- \frac{1}{5} e^{2 x} cos (5 x) - (- \frac{2}{5} \cdot \int e^{2 x} cos (5 x) d x))

Isolate

\int e^{2 x} cos (5 x) d x

= \frac{5 e ^{2 x} sin ( 5 x )}{29} + \frac{2 e ^{2 x} cos ( 5 x )}{29}

Add a constant to the solution

= \frac{5 e ^{2 x} sin ( 5 x )}{29} + \frac{2 e ^{2 x} cos ( 5 x )}{29} + C

\frac{d y}{d x} + y = cos (x)

\int (5 sin (2 θ))^{2} d θ

\int \frac{1}{x - 8 x + 15} d x

\int (x^{2} + 3 x) d x

\frac{x}{x + 2}

What is the integral of e^{2x}cos(5x) ?
The integral of e^{2x}cos(5x) is (5e^{2x}sin(5x))/(29)+(2e^{2x}cos(5x))/(29)+C