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

The integral of e^{-inpix} is 1/(pin)sin(pinx)+i 1/(pin)cos(pinx)+C

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integral of e^{-inpix}

\int e^{- inπ x} d x

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

Solution steps

\int e^{- inπ x} d x

Simplify

e^{- inπ x} : cos (πn x) - i sin (πn x)

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of e^{-inpix} ?
The integral of e^{-inpix} is 1/(pin)sin(pinx)+i 1/(pin)cos(pinx)+C