What is the integral of e^{i2x} ?

The integral of e^{i2x} is 1/2 sin(2x)-i 1/2 cos(2x)+C

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integral of e^{i2x}

\int e^{i 2 x} d x

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

Solution steps

\int e^{i 2 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

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What is the integral of e^{i2x} ?
The integral of e^{i2x} is 1/2 sin(2x)-i 1/2 cos(2x)+C