What is the integral from 0 to (2pi)/3 of e^{2it} ?

The integral from 0 to (2pi)/3 of e^{2it} is -(sqrt(3))/4+i 3/4

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integral from 0 to (2pi)/3 of e^{2it}

\int_{0}^{\frac{2 π}{3}} e^{2 i t} d t

- \frac{3}{4} + i \frac{3}{4}

Solution steps

\int_{0}^{\frac{2 π}{3}} e^{2 i t} d t

Apply imaginary number rule:

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

= \int_{0}^{\frac{2 π}{3}} cos (2 t) + i sin (2 t) d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{\frac{2 π}{3}} cos (2 t) d t + \int_{0}^{\frac{2 π}{3}} i sin (2 t) d t

\int_{0}^{\frac{2 π}{3}} cos (2 t) d t = - \frac{3}{4}

\int_{0}^{\frac{2 π}{3}} i sin (2 t) d t = \frac{3 i}{4}

= - \frac{3}{4} + \frac{3 i}{4}

Simplify

- \frac{3}{4} + \frac{3 i}{4} : - \frac{3}{4} + i \frac{3}{4}

= - \frac{3}{4} + i \frac{3}{4}

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What is the integral from 0 to (2pi)/3 of e^{2it} ?
The integral from 0 to (2pi)/3 of e^{2it} is -(sqrt(3))/4+i 3/4