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

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

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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

Simplify

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

= \int cos (wt) - i sin (wt) 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 (wt) d t - \int i sin (wt) d t

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

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

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

Simplify

\frac{1}{w} sin (wt) - (- i \frac{1}{w} cos (wt)) : \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} (- 3 x^{2} + 3)

x \to 0 lim (\frac{csc ( 3 x )}{cot ( x )})

\frac{\partial}{\partial y} (y^{6} cos (5 x))

\int_{1}^{2} \frac{7 x - 2}{3 x ^{2} - 2 x} d x

f (x) = (4 x + 3) (- x + 1), at x = 4

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