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

Apply u-substitution

= \int - \frac{i e ^{u}}{w} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{i}{w} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - \frac{i}{w} e^{u}

Substitute back

u = w i x

= - \frac{i}{w} e^{w i x}

Simplify

- \frac{i}{w} e^{w i 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

\int \frac{1}{( x 1 - x ^{2} )} d x

\int 4 x^{2} d x

\int x^{3} + x^{2} d x

\int \frac{x ^{9}}{x ^{10} - 1} d x

\int x^{3} - 4 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