What is the integral of e^{itx} ?

The integral of e^{itx} is 1/t sin(tx)-i 1/t cos(tx)+C

integral of e^{itx}

\int e^{i t x} d x

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

Solution steps

\int e^{i t x} d x

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = t i x

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

Simplify

- \frac{i}{t} e^{t i x} : \frac{1}{t} sin (t x) - i \frac{1}{t} cos (t x)

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

Add a constant to the solution

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

\int (ln (x))^{4} d x

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\frac{\partial}{\partial y} (y e^{x^{2} - y})

\int x (sin (x^{\frac{3}{2}} - 1))^{2} d x

\frac{d}{d x} ((2 x^{2} - 7 x + 6)^{2})

What is the integral of e^{itx} ?
The integral of e^{itx} is 1/t sin(tx)-i 1/t cos(tx)+C