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

The integral of e^{-inwt} is 1/(wn)sin(wnt)+i 1/(wn)cos(wnt)+C

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integral of e^{-inwt}

\int e^{- in wt} d t

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

Solution steps

\int e^{- in wt} d t

Simplify

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (- 3 \cdot e^{x^{2} + x - 1})

\frac{\partial}{\partial x} (- ln (1 + e^{- x}))

f (w) = (6 x - 8)^{2}

f (x) = 7 arctan (x - 1 + x^{2})

f (x) = x^{x^{x}}

What is the integral of e^{-inwt} ?
The integral of e^{-inwt} is 1/(wn)sin(wnt)+i 1/(wn)cos(wnt)+C