What is the integral of e^{i2t} ?

The integral of e^{i2t} is 1/2 sin(2t)-i 1/2 cos(2t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of e^{i2t}

\int e^{i 2 t} d t

\frac{1}{2} sin (2 t) - i \frac{1}{2} cos (2 t) + C

Solution steps

\int e^{i 2 t} d t

Apply imaginary number rule:

e^{ia} = cos (a) + i sin (a)

= \int cos (2 t) + i sin (2 t) d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int cos (2 t) d t + \int i sin (2 t) d t

\int cos (2 t) d t = \frac{1}{2} sin (2 t)

\int i sin (2 t) d t = - i \frac{1}{2} cos (2 t)

= \frac{1}{2} sin (2 t) - i \frac{1}{2} cos (2 t)

Add a constant to the solution

= \frac{1}{2} sin (2 t) - i \frac{1}{2} cos (2 t) + C

What is the integral of e^{i2t} ?
The integral of e^{i2t} is 1/2 sin(2t)-i 1/2 cos(2t)+C