What is the integral from 0 to pi/6 of e^{i2t} ?

The integral from 0 to pi/6 of e^{i2t} is (sqrt(3))/4+i 1/4

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integral from 0 to pi/6 of e^{i2t}

\int_{0}^{\frac{π}{6}} e^{i 2 t} d t

\frac{3}{4} + i \frac{1}{4}

Solution steps

\int_{0}^{\frac{π}{6}} e^{i 2 t} d t

Apply imaginary number rule:

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

= \int_{0}^{\frac{π}{6}} 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{π}{6}} cos (2 t) d t + \int_{0}^{\frac{π}{6}} i sin (2 t) d t

\int_{0}^{\frac{π}{6}} cos (2 t) d t = \frac{3}{4}

\int_{0}^{\frac{π}{6}} i sin (2 t) d t = \frac{i}{4}

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

Simplify

\frac{3}{4} + \frac{i}{4} : \frac{3}{4} + i \frac{1}{4}

= \frac{3}{4} + i \frac{1}{4}

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What is the integral from 0 to pi/6 of e^{i2t} ?
The integral from 0 to pi/6 of e^{i2t} is (sqrt(3))/4+i 1/4