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

The integral of e^{-int} is 1/n sin(nt)+i 1/n cos(nt)+C

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integral of e^{-int}

\int e^{- in t} d t

\frac{1}{n} sin (n t) + i \frac{1}{n} cos (n t) + C

Solution steps

\int e^{- in t} d t

Simplify

e^{- in t} : cos (n t) - i sin (n t)

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

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

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

= \frac{1}{n} sin (n t) - (- i \frac{1}{n} cos (n t))

Simplify

\frac{1}{n} sin (n t) - (- i \frac{1}{n} cos (n t)) : \frac{1}{n} sin (n t) + i \frac{1}{n} cos (n t)

= \frac{1}{n} sin (n t) + i \frac{1}{n} cos (n t)

Add a constant to the solution

= \frac{1}{n} sin (n t) + i \frac{1}{n} cos (n t) + C

What is the integral of e^{-int} ?
The integral of e^{-int} is 1/n sin(nt)+i 1/n cos(nt)+C