What is the integral of e^{ix} ?

The integral of e^{ix} is sin(x)-icos(x)+C

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integral of e^{ix}

\int e^{i x} d x

sin (x) - i cos (x) + C

Solution steps

\int e^{i x} d x

Apply u-substitution

: \int - i e^{u} d u

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

Take the constant out:

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

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

Use the common integral:

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

= - i e^{u}

Substitute back

u = i x

= - i e^{i x}

Simplify

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

= sin (x) - i cos (x)

Add a constant to the solution

= sin (x) - i cos (x) + C

\frac{1}{sin ( x )}

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\int (z + 3)^{4} d z

(3 x - 2)^{3} (8 x^{2} + 7)^{3}