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

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