What is the integral of e^{-ikx} ?

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

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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

Simplify

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

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

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

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

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

Simplify

\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)

= \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

\int \frac{1}{e ^{2 x} - 9} d x

\int \frac{1}{x} \frac{x + 1}{x} d x

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