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

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

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integral of e^{-inx}

\int e^{- in x} d x

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

Solution steps

\int e^{- in x} d x

Simplify

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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