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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

\int (x cos (y)) d y

\frac{\partial}{\partial x} (y - x)

\int \frac{sin ( 2 x )}{cos ( x )} d x

\frac{d y}{d x} = 5 x

\int_{- 2}^{2} x^{2} + 1 d x

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