What is the integral of e^{-2ix} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{-2ix}

\int e^{- 2 i x} d x

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

Solution steps

\int e^{- 2 i x} d x

Simplify

e^{- 2 i x} : cos (2 x) - i sin (2 x)

= \int cos (2 x) - i sin (2 x) d x

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 x) d x - \int i sin (2 x) d x

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

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

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

Simplify

\frac{1}{2} sin (2 x) - (- i \frac{1}{2} cos (2 x)) : \frac{1}{2} sin (2 x) + i \frac{1}{2} cos (2 x)

= \frac{1}{2} sin (2 x) + i \frac{1}{2} cos (2 x)

Add a constant to the solution

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

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

\int \frac{( 3 + x )}{x} d x

\int \frac{1}{2 e ^{x} + 3} d x

\int (3 + 7 x^{2})^{9} 0.5 x d x

\int \frac{ln ( k )}{k} d k

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