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

The integral of e^{-iwx} is 1/w sin(wx)+i 1/w cos(wx)+C

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integral of e^{-iwx}

\int e^{- i w x} d x

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

Solution steps

\int e^{- i w x} d x

Simplify

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

(x - 1) y^{^{'}} + y (x - 2) = 0

f (x) = sin (2) x^{2}

x^{^{''}} - 5 x^{^{'}} + 6 x = e^{2 t}

y = (x^{2} - 6 x + 10) 25 - x^{2}

\int 13 tan^{4} (x se) c^{6} x d x

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