What is the integral of e^{2ix} ?

The integral of e^{2ix} is 1/2 sin(2x)-i 1/2 cos(2x)+C

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integral of e^{2ix}

\int e^{2 i x} d x

\frac{1}{2} sin (2 x) - i \frac{1}{2} cos (2 x) + C

Solution steps

\int e^{2 i x} d x

Apply imaginary number rule:

e^{ia} = cos (a) + i sin (a)

= \int cos (2 x) + i sin (2 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 (2 x) d x + \int i sin (2 x) d x

\int cos (2 x) d x = \frac{1}{2} sin (2 x)

\int i sin (2 x) d x = - i \frac{1}{2} cos (2 x)

= \frac{1}{2} sin (2 x) - i \frac{1}{2} cos (2 x)

Add a constant to the solution

= \frac{1}{2} sin (2 x) - i \frac{1}{2} cos (2 x) + C

\int_{2}^{3} (y^{2} - 2 y + 1) d y

\frac{d}{d t} (\frac{3 t}{1 + t ^{3}})

x \to 0 + lim (ln (4 sin (x)))

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

\int \frac{1}{5 - 2 x} d x

What is the integral of e^{2ix} ?
The integral of e^{2ix} is 1/2 sin(2x)-i 1/2 cos(2x)+C