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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of cos(2x)e^x

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

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

Solution steps

\int cos (2 x) e^{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

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

= 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

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