What is the integral of e^{ikx} ?

The integral of e^{ikx} is 1/k sin(kx)-i 1/k cos(kx)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of e^{ikx}

\int e^{ik x} d x

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

Solution steps

\int e^{ik x} d x

Apply u-substitution

= \int - \frac{i e ^{u}}{k} d u

Take the constant out:

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

= - \frac{i}{k} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - \frac{i}{k} e^{u}

Substitute back

u = ik x

= - \frac{i}{k} e^{ik x}

Simplify

- \frac{i}{k} e^{ik x} : \frac{1}{k} sin (k x) - i \frac{1}{k} cos (k x)

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

Add a constant to the solution

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

What is the integral of e^{ikx} ?
The integral of e^{ikx} is 1/k sin(kx)-i 1/k cos(kx)+C