What is the integral of e^{iwt} ?

The integral of e^{iwt} is 1/w sin(wt)-i 1/w cos(wt)+C

integral of e^{iwt}

\int e^{i wt} d t

\frac{1}{w} sin (wt) - i \frac{1}{w} cos (wt) + C

Solution steps

\int e^{i wt} d t

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = w i t

= - \frac{i}{w} e^{w i t}

Simplify

- \frac{i}{w} e^{w i t} : \frac{1}{w} sin (wt) - i \frac{1}{w} cos (wt)

= \frac{1}{w} sin (wt) - i \frac{1}{w} cos (wt)

Add a constant to the solution

= \frac{1}{w} sin (wt) - i \frac{1}{w} cos (wt) + C

\frac{d}{d x} (1 - (\frac{x + b + c}{( 3 a )}))

\int_{3}^{5} \frac{1}{x + 1} d x

\frac{d}{d x} (\frac{1}{2} e^{- \frac{x}{2}})

y^{^{'}} = \frac{cos ( 14 x )}{e ^{14 y}}

\frac{d}{d x} (\frac{( x )}{x ^{2} + 1})

What is the integral of e^{iwt} ?
The integral of e^{iwt} is 1/w sin(wt)-i 1/w cos(wt)+C